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Question 1 of 100
In the diagram below, ACF is a angled triangle, DEGH is a square and ABDH is a rhombus. Give that AB = BH, find$\angle$AFC.
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15$^\circ$ $\rightarrow$ $360 – 60 – 60 – 90 = 150$
$180 – 150 = 30$
$30 ÷ 2 = 15$
$60 + 15 = 75$
$180 – 90 – 75 = 15$
You are Right
15$^\circ$ $\rightarrow$ $360 – 60 – 60 – 90 = 150$
$180 – 150 = 30$
$30 ÷ 2 = 15$
$60 + 15 = 75$
$180 – 90 – 75 = 15$
In the figure below not drawn to scale, ABDE is a trapezium and ABCE is a parallelogram. Give that $\angle$BAE = 125$^\circ$, find the sum of $\angle$AEC and $\angle$ECD.
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In the diagram below, EF is parallel to GH and JF and KE are straight line. Find$\angle$KGH.
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$180 – 145 = 35$ $\angle$GAH
$\angle$JHG = $\angle$JFE = 68$^\circ$
$180 – 35 – 68 = 77^\circ$
You are Right
$180 – 145 = 35$ $\angle$GAH
$\angle$JHG = $\angle$JFE = 68$^\circ$
$180 – 35 – 68 = 77^\circ$
In the diagram below, not drawn to scale, ABCD is a rhombus, and DEC is an equilateral triangle. If $\angle$DAC = 43 $^\circ$. Find$\angle$CBE.
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17$^\circ$ $\rightarrow$ $60^\circ + 43^\circ + 43^\circ = 146^\circ$
$180^\circ - 146^\circ = 34^\circ$
$34^\circ ÷ 2^\circ = 17^\circ$
You are Right
17$^\circ$ $\rightarrow$ $60^\circ + 43^\circ + 43^\circ = 146^\circ$
$180^\circ - 146^\circ = 34^\circ$
$34^\circ ÷ 2^\circ = 17^\circ$
The figure below is not drawn to scale. PQXS is a parallelogram, QRX and QXY are isosceles triangles, TW // PQ and $\angle$a PUT = 120 $^\circ$. SXR is a straight line.
(a)Find$\angle$a RQX
(b) Find$\angle$a SYQ
(b) 141$^\circ$
(b) 142$^\circ$
(b) 143$^\circ$
(b) 144$^\circ$
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(a) 180$^\circ$ - 102$^\circ$ = 78$^\circ$
$78^\circ \times 2 = 156^\circ$
$\angle$RQX $\rightarrow$ 180$^\circ$ - 156$^\circ$ = 24$^\circ$
(b) 180$^\circ$ - 78$^\circ$ = 102$^\circ$
$(180^\circ - 102^\circ)$ ÷ 2 = 39$^\circ$
$\angle$SYQ $\rightarrow$ 180$^\circ$ - 39$^\circ$ = 141$^\circ$
(b) 141$^\circ$
You are Right
(a) 180$^\circ$ - 102$^\circ$ = 78$^\circ$
$78^\circ \times 2 = 156^\circ$
$\angle$RQX $\rightarrow$ 180$^\circ$ - 156$^\circ$ = 24$^\circ$
(b) 180$^\circ$ - 78$^\circ$ = 102$^\circ$
$(180^\circ - 102^\circ)$ ÷ 2 = 39$^\circ$
$\angle$SYQ $\rightarrow$ 180$^\circ$ - 39$^\circ$ = 141$^\circ$
The figure below shown three overlapping triangle. ABC is an isosceles triangle and AB // FK. $\angle$ ABC = 106$^\circ$, $\angle$ CDH = 18$^\circ$, $\angle$ KFH = 52$^\circ$. and $\angle$ GJH = 40$^\circ$. Find
(a) $\angle$ FHD.
(b) $\angle$FKG.
(b) 10$^\circ$
(b) 15$^\circ$
(b) 20$^\circ$
(b) 25$^\circ$
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(a) $\angle$BEF = $\angle$CBA = $(180^\circ – 106^\circ) ÷ 2 = 37^\circ$
$\angle$BEF = $\angle$DEC = 37$^\circ$
$\angle$CDE = $180^\circ – 37^\circ - 106^\circ = 37^\circ$
$37 + 18 = 55^\circ$
$\angle$FHD = $180^\circ - 52^\circ - 55^\circ = 73^\circ$
(b) $\angle$JGH = $180^\circ - 73^\circ - 40^\circ = 67^\circ$
$\angle$ FGK = 113%
$\angle$FKG = $180^\circ - 113^\circ - 52^\circ = 15^\circ$
(b) 15$^\circ$
You are Right
(a) $\angle$BEF = $\angle$CBA = $(180^\circ – 106^\circ) ÷ 2 = 37^\circ$
$\angle$BEF = $\angle$DEC = 37$^\circ$
$\angle$CDE = $180^\circ – 37^\circ - 106^\circ = 37^\circ$
$37 + 18 = 55^\circ$
$\angle$FHD = $180^\circ - 52^\circ - 55^\circ = 73^\circ$
(b) $\angle$JGH = $180^\circ - 73^\circ - 40^\circ = 67^\circ$
$\angle$ FGK = 113%
$\angle$FKG = $180^\circ - 113^\circ - 52^\circ = 15^\circ$
In the figure below, not drawn to scale, JKLM is a parallelogram. MKN is a straight line find the value of $\angle$LKN.
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In the diagram below, ABCD is a trapezium and AEF is an equilateral triangle. Find
(a) $\angle$BAD.
(b) $\angle$DFE.
(b) 5$^\circ$
(b) 10$^\circ$
(b) 15$^\circ$
(b) 20$^\circ$
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(a) 80$^\circ$
(b) $180^\circ – 80^\circ - 60^\circ = 40^\circ$
$60^\circ – 40 = 20^\circ$
(b) 20$^\circ$
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(a) 80$^\circ$
(b) $180^\circ – 80^\circ - 60^\circ = 40^\circ$
$60^\circ – 40 = 20^\circ$
In the figure below, XYZ is an isosceles triangle. XY = YZ and AB// XY. AWY and AVB are straight line. Find $\angle$ t.
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$\angle$AYZ = $180^\circ – 20^\circ – 124^\circ = 36^\circ$
$\angle$YXZ + $\angle$XYZ = $180^\circ – 20^\circ – 36^\circ = 124^\circ$
$\angle$YXZ = $180^\circ - 62^\circ - 20^\circ = 98^\circ$
You are Right
$\angle$AYZ = $180^\circ – 20^\circ – 124^\circ = 36^\circ$
$\angle$YXZ + $\angle$XYZ = $180^\circ – 20^\circ – 36^\circ = 124^\circ$
$\angle$YXZ = $180^\circ - 62^\circ - 20^\circ = 98^\circ$
In the figure below, PQRS is a rectangle and QTUR is a square. PQT and SRU are straight line. Find $\angle$ SQU.
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In the figure below, ABCD is a rhombus and CDEF is a parallelogram. $\angle$ADE is 150$^\circ$ and $\angle$CFE is 115$^\circ$. Find$\angle$x.
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$360 – 150 – 115 = 95$
$360 – 95 – 95 = 170$
$170 ÷ 2 = 85$
$\angle$x = $85 ÷ 2 = 42.5^\circ$
You are Right
$360 – 150 – 115 = 95$
$360 – 95 – 95 = 170$
$170 ÷ 2 = 85$
$\angle$x = $85 ÷ 2 = 42.5^\circ$
In the figure below (not drawn to scale), AOB is a straight line and$\angle$EOC is a right angle, $\angle$DOE =50$^\circ$ and $\angle$BOC =39$^\circ$. Find $\angle$AOD.
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79$^\circ$ $\rightarrow$ $\angle$ $EOB = 90^\circ 39^\circ = 51^\circ$
$\angle$ $AOC = 180^\circ - 39^\circ = 141^\circ$
$\angle$ $AOD = 360^\circ (180^\circ+50^\circ+51^\circ) = 360^\circ - 281^\circ = 79^\circ$
You are Right
79$^\circ$ $\rightarrow$ $\angle$ $EOB = 90^\circ 39^\circ = 51^\circ$
$\angle$ $AOC = 180^\circ - 39^\circ = 141^\circ$
$\angle$ $AOD = 360^\circ (180^\circ+50^\circ+51^\circ) = 360^\circ - 281^\circ = 79^\circ$
The figure not drawn to scale, is made up of a triangle ABC and two rhombuses, PSRQ and PTCU. $\angle$ TPU = 79$^\circ$. Find $\angle$ d + $\angle$ e + $\angle$ f + $\angle$ g.
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$\angle$d + $\angle$e = $180^\circ = 79^\circ = 101^\circ$
$\angle$d + $\angle$e + $\angle$f + $\angle$g = $101^\circ + 202^\circ = 303^\circ$
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$\angle$d + $\angle$e = $180^\circ = 79^\circ = 101^\circ$
$\angle$d + $\angle$e + $\angle$f + $\angle$g = $101^\circ + 202^\circ = 303^\circ$
In the figure below, ABC is an equilateral triangle and CD and BD are straight line. DC is parallel to AB. $\angle$ CDE is 36 $^\circ$. Find $\angle$ DEA.
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In the figure below not scale, XYZ is an isosceles triangle where XZ =ZY. XZW is a straight line. Three angles are labelled as a, b and c. Which of the following statement is true?
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In the figure below not drawn to scale, UTS is a straight line. PQ is parallel to RS and QT = QS. Give that $\angle$TSQ = 30$^\circ$, Find$\angle$RST.
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In the figure below, not drawn to scale, ABCD, HKJC and BGFE are squares, $\angle$ BKJ = 50 $^\circ$. and $\angle$ CBE = 70 $^\circ$.Find$\angle$ AHC.
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$\angle$ HKB $\rightarrow$ $90^\circ - 50^\circ = 40^\circ$
$\angle$KBH $\rightarrow$ $360^\circ - 180^\circ - 70^\circ = 110^\circ$
$\angle$BHK $\rightarrow$ $180^\circ - 110^\circ - 40^\circ = 30^\circ$
$\angle$BHC $\rightarrow$ $90^\circ - 30^\circ = 60^\circ$
$\angle$AHC $\rightarrow$ $180^\circ - 60^\circ = 120^\circ$
You are Right
$\angle$ HKB $\rightarrow$ $90^\circ - 50^\circ = 40^\circ$
$\angle$KBH $\rightarrow$ $360^\circ - 180^\circ - 70^\circ = 110^\circ$
$\angle$BHK $\rightarrow$ $180^\circ - 110^\circ - 40^\circ = 30^\circ$
$\angle$BHC $\rightarrow$ $90^\circ - 30^\circ = 60^\circ$
$\angle$AHC $\rightarrow$ $180^\circ - 60^\circ = 120^\circ$
The figure shown below is not drawn to scale. AB // CD and AD // BC. Find $\angle$m.
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In the figure, PQST is a parallelogram and PQR is an isosceles triangle. PR = QR, $\angle$TPQ = 136$^\circ$ and $\angle$SQR = 18$^\circ$.
Find $\angle$PRQ
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In the figure below, not drawn to scale, LPRS is a parallelogram. ML = NL and NQ and MS are straight line. Find $\angle$PQR.
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In the figure below, a rectangular piece of paper is folded at a corner as shown Find $\angle$m.
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In the figure below, not drawn to scale, ABCD is a parallelogram. GED, GHKF and BCF are straight line. $\angle$ DAE = 110$^\circ$, $\angle$ EGH = 60$^\circ$, and $\angle$ KFC = 30 $^\circ$.
(a) Find $\angle$ KCF.
(b) Find $\angle$ AEG.
(b) 120$^\circ$
(b) 130$^\circ$
(b) 140$^\circ$
(b) 150$^\circ$
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(a)$\angle$KCF $\rightarrow$ $180^\circ - 110^\circ = 70^\circ$
(b) $\angle$ FKC $\rightarrow$ $180 - 70^\circ - 30^\circ = 80^\circ$
$\angle$HEG $\rightarrow$ $180^\circ - 80^\circ - 60^\circ = 40^\circ$
$\angle$ AEG $\rightarrow$ $180^\circ - 40^\circ = 140^\circ$
(b) 140$^\circ$
You are Right
(a)$\angle$KCF $\rightarrow$ $180^\circ - 110^\circ = 70^\circ$
(b) $\angle$ FKC $\rightarrow$ $180 - 70^\circ - 30^\circ = 80^\circ$
$\angle$HEG $\rightarrow$ $180^\circ - 80^\circ - 60^\circ = 40^\circ$
$\angle$ AEG $\rightarrow$ $180^\circ - 40^\circ = 140^\circ$
In the figure below, two comers of a triangle piece of paper are folded inward, then outward to form a symmetrical shape as shown in Figure B. $\angle$PQR = 11$^\circ$, and $\angle$PQS = 23$^\circ$. Find$\angle$TQR.
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22$^\circ$ $\rightarrow$ $11^\circ + 23^\circ = 34^\circ$
$34^\circ \times 2 = 68^\circ$
$\angle$ TQR $\rightarrow$ $90^\circ - 58^\circ = 22^\circ$
You are Right
22$^\circ$ $\rightarrow$ $11^\circ + 23^\circ = 34^\circ$
$34^\circ \times 2 = 68^\circ$
$\angle$ TQR $\rightarrow$ $90^\circ - 58^\circ = 22^\circ$
In the figure shown below, ADGJ is a rectangle, GHJK is a rhombus and DEFG is a parallelogram. $\angle$GHJ = 78 $^\circ $, and $\angle$FGH =92 $^\circ $. Find
(a) $\angle$CGD.
(b) $\angle$GFE.
(b) 51$^\circ$
(b) 52$^\circ$
(b) 53$^\circ$
(b) 54$^\circ$
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$\angle$CGJ $\rightarrow (180 – 78) ÷ 2 = 51$
$\angle$CGD $\rightarrow$ 90 – 51 = 39
$\angle$DGF $\rightarrow$ $360 – 92 – 51 – 51 - 39 = 127$
$\angle$GFE $\rightarrow$ 180 – 127 = 53
$\angle$CGD is 39$^\circ$
$\angle$GFE is 53$^\circ$
(b) 53$^\circ$
You are Right
$\angle$CGJ $\rightarrow (180 – 78) ÷ 2 = 51$
$\angle$CGD $\rightarrow$ 90 – 51 = 39
$\angle$DGF $\rightarrow$ $360 – 92 – 51 – 51 - 39 = 127$
$\angle$GFE $\rightarrow$ 180 – 127 = 53
$\angle$CGD is 39$^\circ$
$\angle$GFE is 53$^\circ$
In the figure below, ABD is a triangle and ABC is a straight line. Given that $\angle$y is thrice $\angle$x, find $\angle$x
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In the figure below, not drawn to scale, two rectangles ABHI and GHJK overlap each other as shown . Given that AD // GF and CF//DE. Find
(a) $\angle$p
(b) $\angle$q
(b) 210$^\circ$
(b) 215$^\circ$
(b) 220$^\circ$
(b) 225$^\circ$
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$a 90^\circ - 75^\circ = 15^\circ$
$\angle$$p = 90^\circ – 15 = 75^\circ$
$b 180^\circ - 120^\circ = 60^\circ$
$\angle$$q = 360^\circ - 60^\circ - 85^\circ = 215^\circ$
(b) 215$^\circ$
You are Right
$a 90^\circ - 75^\circ = 15^\circ$
$\angle$$p = 90^\circ – 15 = 75^\circ$
$b 180^\circ - 120^\circ = 60^\circ$
$\angle$$q = 360^\circ - 60^\circ - 85^\circ = 215^\circ$
The figure below is made up of a triangle and a trapezium. Find the sum of all the unknown marked angle in the figure.
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The diagram shown below is not draw to scale. QRST is a rhombus and STUM is a trapezium. Find $\angle$a
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78$^\circ$ $\rightarrow$ $\angle$ RST $\rightarrow$ $180^\circ – (2 \times 28^\circ) = 124^\circ$
$\angle$ VSW $\rightarrow$ $180^\circ – (2 \times 51^\circ) = 78^\circ$
You are Right
78$^\circ$ $\rightarrow$ $\angle$ RST $\rightarrow$ $180^\circ – (2 \times 28^\circ) = 124^\circ$
$\angle$ VSW $\rightarrow$ $180^\circ – (2 \times 51^\circ) = 78^\circ$
The figure below shown four identical quadrants inside a square, ABCD. Find the perimeter of the shaded region. $ (Take \pi= \frac{22}{7})$
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A rectangular piece of paper is folded along the dotted line AC as shown below. Find
(a) $\angle$ DAE
(b) $\angle$ ACE
(b) 61$^\circ$
(b) 63$^\circ$
(b) 65$^\circ$
(b) 67$^\circ$
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(a) $\angle$ $DAE = 90^\circ - 54^\circ = 36^\circ$
(b) 63$^\circ$ $\rightarrow$ $\angle$ $ACE = 180^\circ – (27^\circ+90^\circ) = 63^\circ$
(b) 63$^\circ$
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(a) $\angle$ $DAE = 90^\circ - 54^\circ = 36^\circ$
(b) 63$^\circ$ $\rightarrow$ $\angle$ $ACE = 180^\circ – (27^\circ+90^\circ) = 63^\circ$
The figure below shows an equilateral triangular piece of paper folded along line CX. $\angle$ACB is 22 $^\circ$. Find $\angle$ BCX.
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In the figure below, not drawn to scale, WXYZ is a trapezium. AY, GW and YC are straight line. Find $\angle$YWX
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$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$
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$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$
In the figure below, AOC is a straight line $\angle$AOB = 159$^\circ$ and $\angle$COD = 63$^\circ$. What is the sum of$\angle$ AOD and$\angle$BOC?
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ABC and OBC are isosceles triangles. AB= AC, OB= OC, $\angle$ ABO=20$^\circ$ and $\angle$ BOC=80$^\circ$. Find$\angle$a.
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$(180 – 80) ÷ 2 = 50$
50 + 20 = 70
$70 \times 2 = 140$
180 – 140 = 40$^\circ$
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$(180 – 80) ÷ 2 = 50$
50 + 20 = 70
$70 \times 2 = 140$
180 – 140 = 40$^\circ$
In the figure below which is not dram to scale, EFGH is a parallelogram. O is the center of the circle. Find
(a) $\angle$x
(b) $\angle$y
(b) 90$^\circ$
(b) 100$^\circ$
(b) 120$^\circ$
(b) 140$^\circ$
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(a) 90$^\circ$
(b) $\angle$HFO = $\angle$OEF = $\angle$ OFE = 60$^\circ$
$\angle$HFO = Isosceles
$\angle$ FHO = $\angle$HFO = $(180^\circ - 120^\circ) ÷ 2 = 30^\circ$
$\angle$x = $30^\circ + 60^\circ = 90^\circ$
$\angle$ GFE = $\angle$GHO = $\angle$y = 120$^\circ$
(b) 120$^\circ$
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(a) 90$^\circ$
(b) $\angle$HFO = $\angle$OEF = $\angle$ OFE = 60$^\circ$
$\angle$HFO = Isosceles
$\angle$ FHO = $\angle$HFO = $(180^\circ - 120^\circ) ÷ 2 = 30^\circ$
$\angle$x = $30^\circ + 60^\circ = 90^\circ$
$\angle$ GFE = $\angle$GHO = $\angle$y = 120$^\circ$
The figure below shows a trapezium ABCD with AB // DC, $\angle$ DAB = 68 $^\circ$. And $\angle$ ABC is an isosceles triangle with AB = AD. Find $\angle$BCD.
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30$^\circ$ $\rightarrow$ $\angle$ $ABD = (180^\circ-68^\circ) ÷ 2 = 56^\circ$
$\angle$ ABD = $\angle$ BDC
$\angle$ $BCD = 180^\circ – (94^\circ+56^\circ) = 30^\circ$
You are Right
30$^\circ$ $\rightarrow$ $\angle$ $ABD = (180^\circ-68^\circ) ÷ 2 = 56^\circ$
$\angle$ ABD = $\angle$ BDC
$\angle$ $BCD = 180^\circ – (94^\circ+56^\circ) = 30^\circ$
In the diagram below, ABC and DEF are equilateral triangle. $\angle$BGC is 80$^\circ$. Find $\angle$CHF.
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$180 – 80 – 60 = 40$
$180 – 60 – 40 = 80$
$\angle$CHF = $180 – 80 – 60 = 40^\circ$
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$180 – 80 – 60 = 40$
$180 – 60 – 40 = 80$
$\angle$CHF = $180 – 80 – 60 = 40^\circ$
A rectangular piece of paper was folded as shown below. Find $\angle$a.
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In the figure, CDE and CDF are isosceles triangle. $\angle$DCF is three times as large as $\angle$EDF. Find $\angle$DFE.
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108$^\circ$ $\rightarrow$ $4U + 3U + 3U = 10U$
$1U = 180^\circ ÷ 10 = 18^\circ$
$3U = 18^\circ \times 3 = 54^\circ$
$1U = 18^\circ$
$\angle$ $DEF = 180^\circ – 18^\circ = 54^\circ = 108^\circ$
You are Right
108$^\circ$ $\rightarrow$ $4U + 3U + 3U = 10U$
$1U = 180^\circ ÷ 10 = 18^\circ$
$3U = 18^\circ \times 3 = 54^\circ$
$1U = 18^\circ$
$\angle$ $DEF = 180^\circ – 18^\circ = 54^\circ = 108^\circ$
In the figure, ABCD is a rectangle and CEFG is a rhombus$\angle$EFG = 100. $^\circ$. and $\angle$DCG = 135 $^\circ$. Find$\angle$ BCE.
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The figure below is not drawn to scale. ABCD is a parallelogram and ABE is an isosceles triangle. Find $\angle$p.
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10$^\circ$ $\rightarrow$ $180^\circ - 46^\circ - 46^\circ = 88^\circ$
$\angle$ $98^\circ - 88^\circ = 10^\circ$
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10$^\circ$ $\rightarrow$ $180^\circ - 46^\circ - 46^\circ = 88^\circ$
$\angle$ $98^\circ - 88^\circ = 10^\circ$
In the diagram below, KM = MO. LMNO is a square and KLM is a right- angled triangle.
(a) Find $\angle$KMO.
(b) Find $\angle$MNP.
(b) 10$^\circ$
(b) 15$^\circ$
(b) 20$^\circ$
(b) 25$^\circ$
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(a) $90 ÷ 2 = 45$
$45 \times 2 =90$
$180 – 90 = 90$
It is 90$^\circ$
(b) $180 – 45 – 135 = 0$
$180 – 135 – 20 = 25$
It is 25$^\circ$
(b) 25$^\circ$
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(a) $90 ÷ 2 = 45$
$45 \times 2 =90$
$180 – 90 = 90$
It is 90$^\circ$
(b) $180 – 45 – 135 = 0$
$180 – 135 – 20 = 25$
It is 25$^\circ$
ABCD is a square and AC is the diagonal of the square. DF is a straight line and $\angle$BFE is 123$^\circ$. Find $\angle$CED.
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In the diagram below, ABCD is a square and BCD is an equilateral triangle. CFE is a straight lin. Find $\angle$AEF.
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In the diagram below, ABKJ is a parallelogram, LDEG is a trapezium and line AF is a straight line. Give that LD//JK, $\angle$ABC =65$^\circ$, $\angle$MKH =48$^\circ$, $\angle$LHJ = 54$^\circ$. Find
(a) $\angle$JAK
(b) $\angle$KFE
(b) 130$^\circ$
(b) 132$^\circ$
(b) 134$^\circ$
(b) 136$^\circ$
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(a) 67$^\circ$ $\rightarrow$ $180 – 65 – 48 = 76^\circ$
(b) 132$^\circ$ $\rightarrow$ $\angle$ $KFE = 180 – 48 = 132^\circ$
(b) 132$^\circ$
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(a) 67$^\circ$ $\rightarrow$ $180 – 65 – 48 = 76^\circ$
(b) 132$^\circ$ $\rightarrow$ $\angle$ $KFE = 180 – 48 = 132^\circ$
In the figure, ABCD is a rectangle. $\angle$AFE is 43$^\circ$. Find $\angle$FCD.
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In the figure below not drawn to scale, AOB and COD are straight line. $\angle$AOE = 70$^\circ$ and $\angle$EOF = 220$^\circ$. Find $\angle$x.
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The figure below is not drawn to scale. AB and CD are straight lines. $\angle$AFC =36$^\circ$ and $\angle$ EFD is twice of $\angle$ AEF. Find $\angle$ EFD.
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$\frac{180 - 36}{1 + 2} = 48$ , $48 \times 2 = 96$
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$\frac{180 - 36}{1 + 2} = 48$ , $48 \times 2 = 96$
In the figure below, ST, UV, WX and YZ are straight lines. Which of the following angles, when added up, have the same value as $\angle$UOZ?
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In the figure below. ABFE is a parallelogram and BCDE is trapezium. Given that $\angle$AFE =75 $^\circ$, Find $\angle$y.
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The figure below is not drawn to scale. The ratio of $\angle$x to $\angle$y to $\angle$z is 2 : 3: 7. Find the difference between $\angle$x and$\angle$z.
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The figure is made up of 1 square and 2 isosceles triangle. Find$\angle$ b.
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The figure shown below is not drawn to scale. BCEF is a square. ABF is an equilateral triangle and AED is a straight line. Given that ED = EF, find the sum of $\angle$ECD and $\angle$EDC.
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The figure is not drawn to scale. ABCD is a trapezium. Find $\angle$ADC.
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The figure below is made up of triangles ABC and BCD, and 2 identical circles. Y and Z are the centers of the circle. $\angle$DCB = 86$^\circ$, $\angle$CAB = 58$^\circ$ and $\angle$CDB = 54$^\circ$. Find $\angle$ABD.
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$\angle$CBD $\rightarrow$ $180^\circ - 54^\circ - 86^\circ = 40^\circ$
AC = CB = diameter
$\angle$BAC = $\angle$ABC = 58$^\circ$
$\angle$ABD $\rightarrow$ $58^\circ – 40^\circ = 18^\circ$
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$\angle$CBD $\rightarrow$ $180^\circ - 54^\circ - 86^\circ = 40^\circ$
AC = CB = diameter
$\angle$BAC = $\angle$ABC = 58$^\circ$
$\angle$ABD $\rightarrow$ $58^\circ – 40^\circ = 18^\circ$
In the figure below, a rectangular piece of paper was folded as shown. Find $\angle$ DEB.
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In the figure below, AB and PQ are straight lines. Given that $\angle$BOG= 72 and$\angle$AOQ =108, find$\angle$POG.
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In the figure not scale, SU = SQ, $\angle$ TQP = 34$^\circ$, $\angle$ SQR = 91$^\circ$, and PRQ is parallel to TS. Find $\angle$ TSU.
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In the figure below not, drawn to scale, ABDE is a parallelogram. $\angle$ ACB =70$^\circ$. and $\angle$ BAC = 60 $^\circ$. Find $\angle$ EDC.
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The figure below shows a trapezium ABCD. AD is parallel to BE and BE = BC. $\angle$BEC = 70$^\circ$
(a) Find $\angle$ADC
(b) Find $\angle$ABC.
(b) 100$^\circ$
(b) 105$^\circ$
(b) 110$^\circ$
(b) 115$^\circ$
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(b) 110$^\circ$
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The figure, not down to scale, shows a rhombus ABCD. DB is a straight line. Find $\angle$EDC.
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The figure below is not scale. ABCD is a parallelogram. EFG and JKL are triangle . EG is parallel to JK and EF = FG. $\angle$ JKG = 93$^\circ$, $\angle$ ENL = 70$^\circ$ and $\angle$ EFG = 58$^\circ$.
(a)Find $\angle$ EJK
(b) Find $\angle$ JLK.
(b) 32$^\circ$
(b) 34$^\circ$
(b) 36$^\circ$
(b) 38$^\circ$
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$\angle$ EJK + $\angle$ = $360^\circ - 180^\circ = 180^\circ$
(a) $\angle$ EJK = $180^\circ - 93^\circ = 87^\circ$
$\angle$ EMN = $180^\circ - 61^\circ = 70^\circ = 49^\circ$
$180^\circ - 58^\circ = 122^\circ$
$122 ÷ 2 = 61^\circ$
$\angle$ FGA = $180^\circ - 93^\circ - 61^\circ = 26^\circ$
(b) $\angle$JLK = $180^\circ – 49^\circ - 93^\circ = 38^\circ$
(b) 38$^\circ$
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$\angle$ EJK + $\angle$ = $360^\circ - 180^\circ = 180^\circ$
(a) $\angle$ EJK = $180^\circ - 93^\circ = 87^\circ$
$\angle$ EMN = $180^\circ - 61^\circ = 70^\circ = 49^\circ$
$180^\circ - 58^\circ = 122^\circ$
$122 ÷ 2 = 61^\circ$
$\angle$ FGA = $180^\circ - 93^\circ - 61^\circ = 26^\circ$
(b) $\angle$JLK = $180^\circ – 49^\circ - 93^\circ = 38^\circ$
The figure below is not drawn to scale. ABCD is a trapezium and $\angle$ BDC = $\angle$ ADB. Find $\angle$ CAD.
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120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$
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120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$
In the figure shown below, O is the center of the circle. OS = ST and ROT is a straight line. Find (a) $\angle$ TOS
(b) $\angle$ TRS
10$^\circ$
20$^\circ$
30$^\circ$
40$^\circ$
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(a) 60$^\circ$ $\rightarrow$ $\angle$ $TOS = 180^\circ ÷ 3 = 60^\circ$
(b) 30$^\circ$ $\rightarrow$ $\angle$ $TRS = 180^\circ – 120^\circ ÷ 2 = 30^\circ$
30$^\circ$
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(a) 60$^\circ$ $\rightarrow$ $\angle$ $TOS = 180^\circ ÷ 3 = 60^\circ$
(b) 30$^\circ$ $\rightarrow$ $\angle$ $TRS = 180^\circ – 120^\circ ÷ 2 = 30^\circ$
In the diagram below, ABC is an equilateral triangle. BD and EF are straight line. Find $\angle$y
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AOB is a straight line as shown in the figure. $\angle$a = $\angle$b Find $\angle$a
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32$^\circ$ $\rightarrow$ $64÷ 2 = 32$
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32$^\circ$ $\rightarrow$ $64÷ 2 = 32$
In the figure, CD and MN are straight lines. Which two angles add up to 90$^\circ$ ?
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In the figure below, Find$\angle$X.
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In the figure, AB, CD, EF and GH are straight line and EF is parallel to GH. Which of the following statements is correct?
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In the figure below. ABCD is a rhombus, $\angle$ABC =102$^\circ $ and$\angle$CAE=19$^\circ $. Find $\angle$EAB.
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20$^\circ$ $\rightarrow 180^\circ – 120^\circ = 78^\circ$
$78^\circ ÷ 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
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20$^\circ$ $\rightarrow 180^\circ – 120^\circ = 78^\circ$
$78^\circ ÷ 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
In the diagram below, ABCD is a parallelogram. AC = CE = CF. $\angle$AEC = 55$^\circ $ and $\angle$AFC = 30$^\circ$ AF and CE are straight lines. Find $\angle$ABC.
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$55^\circ - 30^\circ = 25^\circ$
$180^\circ - 55^\circ - 55^\circ = 70^\circ$
$180^\circ - 70^\circ - 30^\circ = 80^\circ$
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$55^\circ - 30^\circ = 25^\circ$
$180^\circ - 55^\circ - 55^\circ = 70^\circ$
$180^\circ - 70^\circ - 30^\circ = 80^\circ$
Which one of the following statements is true?
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The figure below is made up of two identical triangle. TSR and QSR. $\angle$QST = 148$^\circ$. Find $\angle$QSR.
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106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
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106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
In the figure below, BCDG is a trapezium with CB parallel to DG. ABE is an equilateral triangle and AEFG is a square.
(a) Find $\angle$BGE.
(b) Find $\angle$CBE.
(b) 103$^\circ$
(b) 104$^\circ$
(b) 105$^\circ$
(b) 106$^\circ$
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(a) $180 – 60 – 90 = 30$
$30 ÷ 2 = 15$
$45 – 15 = 30$
(b) $60 + 45 = 105^\circ$
(b) 105$^\circ$
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(a) $180 – 60 – 90 = 30$
$30 ÷ 2 = 15$
$45 – 15 = 30$
(b) $60 + 45 = 105^\circ$
The figure below is not drawn to scale. ABCD is a trapezium and CDEF is a parallelogram. Find$\angle$BCF.
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127$^\circ$ $\rightarrow$ $\angle$ $DCF = 180^\circ – 121^\circ = 59^\circ$
$\angle$ $BCD 180^\circ - 112^\circ = 68^\circ$
$\angle$ $BCF = 68^\circ + 59^\circ = 127^\circ$
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127$^\circ$ $\rightarrow$ $\angle$ $DCF = 180^\circ – 121^\circ = 59^\circ$
$\angle$ $BCD 180^\circ - 112^\circ = 68^\circ$
$\angle$ $BCF = 68^\circ + 59^\circ = 127^\circ$
The figure below consists of 3 straight line. What is the value of $\angle$a + $\angle$b +$\angle$c?
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In the diagram below, ABCD is a trapezium and BCE is an equilateral triangle. AB // GF is a straight line. Find $\angle$ABC.
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ABCD is a parallelogram and ADX is an equilateral triangle. $\angle$ABX =50$^\circ$. Find $\angle$ AXB.
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In the figure below, ABEF is a trapezium and BCD is a triangle. ABC is a straight line. $\angle$FDE = $\angle$CDE. Find $\angle$DFE.
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The figure below is not drawn to scale. ACDE is a trapezium. Given the AE = AB, $\angle$EAB =78$^\circ $ and $\angle$ACD is 90 $^\circ $ .
(a) Find $\angle$CDE.
(b) Find $\angle$ABE.
(c) Find $\angle$DEB.
(b) 49$^\circ$
(c) 47$^\circ$
(b) 50$^\circ$
(c) 49$^\circ$
(b) 51$^\circ$
(c) 51$^\circ$
(b) 52$^\circ$
(c) 53$^\circ$
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(a) $180^\circ - 99^\circ = 81^\circ$
(b) $180^\circ - 78^\circ = 102^\circ$
$102^\circ ÷ 2 = 51^\circ$
(c)$\angle$DEB = $\angle$AEB = 51$^\circ$
(b) 51$^\circ$
(c) 51$^\circ$
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(a) $180^\circ - 99^\circ = 81^\circ$
(b) $180^\circ - 78^\circ = 102^\circ$
$102^\circ ÷ 2 = 51^\circ$
(c)$\angle$DEB = $\angle$AEB = 51$^\circ$
The figure below is not drawn to scale. PQR is an isosceles triangle. PRS is a straight line. $\angle$PRS =70$^\circ$ and $\angle$TRS=80$^\circ$. Find the sum of $\angle$a, $\angle$b and $\angle$c.
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155$^\circ$ $\rightarrow$ $\angle$C $\rightarrow$ $180^\circ - 70^\circ ÷ 2 = 55^\circ$
$\angle A + \angle B \rightarrow 180^\circ - 806^\circ = 100^\circ$
Total $100^\circ + 55^\circ = 155^\circ$
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155$^\circ$ $\rightarrow$ $\angle$C $\rightarrow$ $180^\circ - 70^\circ ÷ 2 = 55^\circ$
$\angle A + \angle B \rightarrow 180^\circ - 806^\circ = 100^\circ$
Total $100^\circ + 55^\circ = 155^\circ$
In the figure below, ABC is a triangle. CF and CE are straight line. AD = DB, $\angle$ ADE = 130 $^\circ$. and $\angle$ ABF = 155$^\circ$. Find $\angle$ X.
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$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
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$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
In the figure below, GBEF is a trapezium and BCDE is a parallelogram. ABC and AGF are straight line $\angle$ BEC is five time s of $\angle$CBE. $\angle$FEB is a right angle and $\angle$FAB = 76$^\circ $ Find
(a) $\angle$FED.
(b) $\angle$AFE.
(b) 42$^\circ$
(b) 44$^\circ$
(b) 46$^\circ$
(b) 48$^\circ$
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(a) 120$^\circ$ $\rightarrow$ $\angle$CBE $\rightarrow$ $180^\circ ÷ 6 = 30^\circ$
$\angle$BED $\rightarrow$ $30^\circ \times 5 = 150^\circ$
$\angle$FED $\rightarrow$ $360^\circ - 150^\circ - 90^\circ = 120^\circ$
(b) 44$^\circ$ $\rightarrow$ $\angle$ABG $\rightarrow$ $180^\circ - 90^\circ - 30^\circ = 60^\circ$
$\angle$AGB $\rightarrow$ $180^\circ - 76^\circ - 60^\circ = 44^\circ$
$\angle$BGF $\rightarrow$ $180^\circ - 44^\circ = 136^\circ$
$\angle$AFE $\rightarrow$ $180^\circ - 136^\circ = 44^\circ$
(b) 44$^\circ$
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(a) 120$^\circ$ $\rightarrow$ $\angle$CBE $\rightarrow$ $180^\circ ÷ 6 = 30^\circ$
$\angle$BED $\rightarrow$ $30^\circ \times 5 = 150^\circ$
$\angle$FED $\rightarrow$ $360^\circ - 150^\circ - 90^\circ = 120^\circ$
(b) 44$^\circ$ $\rightarrow$ $\angle$ABG $\rightarrow$ $180^\circ - 90^\circ - 30^\circ = 60^\circ$
$\angle$AGB $\rightarrow$ $180^\circ - 76^\circ - 60^\circ = 44^\circ$
$\angle$BGF $\rightarrow$ $180^\circ - 44^\circ = 136^\circ$
$\angle$AFE $\rightarrow$ $180^\circ - 136^\circ = 44^\circ$
The figure below is made up of three triangle. ABC is an equilateral triangle, BCD is a right-angled triangle and DEF is an isosceles triangle. $\angle$EFD =68$^\circ$. Find the sum of $\angle$CDE and $\angle$ABD.
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194$^\circ$ $\rightarrow$ $180 – 68 – 68 = 44$
$180 – 90 = 90$
$\angle$CDE + $\angle$ABD $\rightarrow$ $90 + 44 + 60 = 194$
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194$^\circ$ $\rightarrow$ $180 – 68 – 68 = 44$
$180 – 90 = 90$
$\angle$CDE + $\angle$ABD $\rightarrow$ $90 + 44 + 60 = 194$
In the figure below, a rectangular piece is folded along AD as shown. Given that EFBC is a straight line , Find
(a) $\angle$EDF
(b) $\angle$DFC
(b) 130$^\circ$
(b) 140$^\circ$
(b) 150$^\circ$
(b) 160$^\circ$
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(a)50$^\circ$
(b) $140 – 180 – 70 – 70 = 40$
$180 – 40 – 90 = 50$
$180 – 5 = 130$
$180 – 90 – 70 = 20$
$90 – 20 – 20 = 50$
$180 – 50 – 90 = 40$
$180 – 40 = 140$
(b) 140$^\circ$
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(a)50$^\circ$
(b) $140 – 180 – 70 – 70 = 40$
$180 – 40 – 90 = 50$
$180 – 5 = 130$
$180 – 90 – 70 = 20$
$90 – 20 – 20 = 50$
$180 – 50 – 90 = 40$
$180 – 40 = 140$
The figure below is not drawn to scale. ABCD is a rhombus. CDE is an isosceles triangle. BCE is a straight line. CE = DE and $\angle$CED = 98$^\circ$. Find $\angle$x.
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In the diagram below , ABCD is a parallelogram and CDEF is a rectangle. Find $\angle$BCF.
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The figure below shown an equilateral triangle XYZ and a straight line AB. Find$\angle$p
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Study the figure below. Find $\angle$k
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The figure below , not drawn to scalel shown a parallelogram ABCD. $\angle$AFG =81$^\circ $, $\angle$BAG =22$^\circ $ and$\angle$ADC =123$^\circ $. Find $\angle$ABC.
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42$^\circ$ $\rightarrow 180 – 123 = 57$
$57 – 22 = 35$
$180 – 35 – 81 = 64$
$180 – 64 = 116$
$180 – 116 – 22 = 42$
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42$^\circ$ $\rightarrow 180 – 123 = 57$
$57 – 22 = 35$
$180 – 35 – 81 = 64$
$180 – 64 = 116$
$180 – 116 – 22 = 42$
The figure below shows a cube. Give that A, B and c are points at the corners of the cube, what is $\angle$ ABC?
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AC = AB = BC
Triangle ABC is an Equilateral Triangle, $\angle$ ABC = 60$^\circ$
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AC = AB = BC
Triangle ABC is an Equilateral Triangle, $\angle$ ABC = 60$^\circ$
In the figure below, ABCD is trapezium and BDE is an isosceles triangle, with DB = EB. Find $\angle$CDE.
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$180^\circ - 50^\circ - 50^\circ = 80^\circ$
$110^\circ - 80^\circ = 30^\circ $
$50^\circ - 30^\circ = 20^\circ$
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$180^\circ - 50^\circ - 50^\circ = 80^\circ$
$110^\circ - 80^\circ = 30^\circ $
$50^\circ - 30^\circ = 20^\circ$
The figure below not drawn to scale shows a parallelogram BCDF and an isosceles triangle ABC. Give that $\angle$FBC = 72 $^\circ$, Find $\angle$ACD.
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In the figure below, not scale, BDEG is a square and BCD is an isosceles triangle. ABC is a straight line. BF//CD and$\angle$ABG = 80$^\circ$.
(a) Find $\angle$BDC.
(b) Find $\angle$BFE.
(b) 75$^\circ$
(b) 85$^\circ$
(b) 95$^\circ$
(b) 105$^\circ$
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(a) $\angle$b $\rightarrow$ $180^\circ - 90^\circ - 80^\circ = 10^\circ$
$\angle$BDC $\rightarrow$ $(180^\circ - 10^\circ) ÷ 2 = 85^\circ$
(b) $\angle$BDC = $\angle$FBD alternate angles in parallel lines
$\angle$FBD = 85$^\circ$
$\angle$BFE = $180^\circ - 85^\circ = 95^\circ$
(b) 95$^\circ$
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(a) $\angle$b $\rightarrow$ $180^\circ - 90^\circ - 80^\circ = 10^\circ$
$\angle$BDC $\rightarrow$ $(180^\circ - 10^\circ) ÷ 2 = 85^\circ$
(b) $\angle$BDC = $\angle$FBD alternate angles in parallel lines
$\angle$FBD = 85$^\circ$
$\angle$BFE = $180^\circ - 85^\circ = 95^\circ$
The figure below, not drawn to scale. Is a circle with O as the center. TS and RW are straight line. The ratio of $\angle$RSO to $\angle$OSW is 2 : 1. $\angle$SWO = 73$^\circ$. Find $\angle$TRS.
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$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
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$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
In the figure, find the sum of $\angle$a, $\angle$b, $\angle$c and $\angle$d.
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$180 – 65 = 115$
$180 ÷ 3 \times 2 = 120$
$120 + 115 = 235$
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$180 – 65 = 115$
$180 ÷ 3 \times 2 = 120$
$120 + 115 = 235$
In the figure below, AB and BC are two sides of a parallelogram ABCD. Measure and with down the value of $\angle$ABC.
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Ali and Tim were standing in a school field as shown in the grid below. Ali was facing south-east and Tim was facing north. How many degrees clockwise must Ali and Tim both turn in order to face each other?
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In the figure, not drawn to scale, PQRS is a parallelogram and TUR is a triangle. If PT = PU, Find
(a) $\angle$TUR
(b) $\angle$URQ
(b) 11$^\circ$
(b) 13$^\circ$
(b) 15$^\circ$
(b) 17$^\circ$
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(a) 50$^\circ$ $\rightarrow$ $180^\circ – (65^\circ \times 2) = 50^\circ$
(b) 13$^\circ$ $\rightarrow$ $\angle$ QUR $\rightarrow$ $180^\circ - 37^\circ - 50^\circ = 93^\circ$
$\angle$ TPU $\rightarrow$ $180^\circ - 37^\circ - 37^\circ = 106^\circ$
$\angle$ UQR $\rightarrow$ $180^\circ - 106^\circ = 74^\circ$
$\angle$ URQ $\rightarrow$ $180^\circ - 93^\circ - 74^\circ = 13^\circ$
(b) 13$^\circ$
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(a) 50$^\circ$ $\rightarrow$ $180^\circ – (65^\circ \times 2) = 50^\circ$
(b) 13$^\circ$ $\rightarrow$ $\angle$ QUR $\rightarrow$ $180^\circ - 37^\circ - 50^\circ = 93^\circ$
$\angle$ TPU $\rightarrow$ $180^\circ - 37^\circ - 37^\circ = 106^\circ$
$\angle$ UQR $\rightarrow$ $180^\circ - 106^\circ = 74^\circ$
$\angle$ URQ $\rightarrow$ $180^\circ - 93^\circ - 74^\circ = 13^\circ$
The figure below is not drawn to scale. ABCD is a square. CXY is a triangle. $\angle$DXY = 180$^\circ$. and $\angle$BCY = 24$^\circ$. Find $\angle$y.
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