- Numbers
- What does the number stand for?
- Nearest number
- Which digit is in the place?
- Round off to the nearest number
- Number divide
- Find the fraction nearest to 1
- Sort fraction numbers
- Find the largest or smallest fraction
- Fraction to decimal
- Convert decimal to a mixed fraction.
- Fraction - Addition
- Fraction - Substraction
- Fraction - Multiply
- Fraction - Divide
- List all the common factors of two integers
- Chicken and Rabbit problem
- Assumption Method
- Measurement
- Data Analysis
- Geometry
- Speed
- Others
- P5 Maths
- P4 Maths
- Tap to switch levels
- Numbers
- Measurement
- Data Analysis
- Geometry
- Speed
- Others
- P5 Maths
- P4 Maths
SELECT A SUBJECT
- Numbers
Question 1 of 100
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.
Sorry. Please check the correct answer below.
194$^\circ$ $\rightarrow$ $180 – 68 – 68 = 44$
$180 – 90 = 90$
$\angle$CDE + $\angle$ABD $\rightarrow$ $90 + 44 + 60 = 194$
You are Right
194$^\circ$ $\rightarrow$ $180 – 68 – 68 = 44$
$180 – 90 = 90$
$\angle$CDE + $\angle$ABD $\rightarrow$ $90 + 44 + 60 = 194$
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.
Sorry. Please check the correct answer below.
You are Right
42$^\circ$ $\rightarrow 180 – 123 = 57$
$57 – 22 = 35$
$180 – 35 – 81 = 64$
$180 – 64 = 116$
$180 – 116 – 22 = 42$
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$^\circ$
Sorry. Please check the correct answer below.
$\angle$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$^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(b) 110$^\circ$
You are Right
Study the figure below. Find $\angle$k
Sorry. Please check the correct answer below.
You are Right
The diagram shown below is not draw to scale. QRST is a rhombus and STUM is a trapezium. Find $\angle$a
Sorry. Please check the correct answer below.
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 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.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(a) 80$^\circ$
(b) $180^\circ – 80^\circ - 60^\circ = 40^\circ$
$60^\circ – 40 = 20^\circ$
(b) 20$^\circ$
You are Right
(a) 80$^\circ$
(b) $180^\circ – 80^\circ - 60^\circ = 40^\circ$
$60^\circ – 40 = 20^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
$\angle$ 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$^\circ$
You are Right
$\angle$ 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 = $180^\circ – 49^\circ - 93^\circ = 38^\circ$
In the figure below. ABFE is a parallelogram and BCDE is trapezium. Given that $\angle$AFE =75 $^\circ$, Find $\angle$y.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(a) 60$^\circ$ $\rightarrow$ $\angle$ $TOS = 180^\circ ÷ 3 = 60^\circ$
(b) 30$^\circ$ $\rightarrow$ $\angle$ $TRS = 180^\circ – 120^\circ ÷ 2 = 30^\circ$
The figure below is not drawn to scale. ABCD is a trapezium and CDEF is a parallelogram. Find$\angle$BCF.
Sorry. Please check the correct answer below.
127$^\circ$ $\rightarrow$ $\angle$ $DCF = 180^\circ – 121^\circ = 59^\circ$
$\angle$ $BCD 180^\circ - 112^\circ = 68^\circ$
$\angle$ $BCF = 68^\circ + 59^\circ = 127^\circ$
You are Right
127$^\circ$ $\rightarrow$ $\angle$ $DCF = 180^\circ – 121^\circ = 59^\circ$
$\angle$ $BCD 180^\circ - 112^\circ = 68^\circ$
$\angle$ $BCF = 68^\circ + 59^\circ = 127^\circ$
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?
Sorry. Please check the correct answer below.
You are Right
The figure, not down to scale, shows a rhombus ABCD. DB is a straight line. Find $\angle$EDC.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
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, 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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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 $\rightarrow$ $180^\circ - 136^\circ = 44^\circ$
In the diagram below , ABCD is a parallelogram and CDEF is a rectangle. Find $\angle$BCF.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(a)50$^\circ$
(b) $140 – 180 – 70 – 70 = 40$
$180 – 40 – 90 = 50$
$180 – 5 = 130$
$180 – 90 – 70 = 20$
$90 – 20 – 20 = 50$
$180 – 50 – 90 = 40$
$180 – 40 = 140$
ABCD is a parallelogram and ADX is an equilateral triangle. $\angle$ABX =50$^\circ$. Find $\angle$ AXB.
Sorry. Please check the correct answer below.
You are Right
In the diagram below, ABC and DEF are equilateral triangle. $\angle$BGC is 80$^\circ$. Find $\angle$CHF.
Sorry. Please check the correct answer below.
$180 – 80 – 60 = 40$
$180 – 60 – 40 = 80$
$\angle$CHF = $180 – 80 – 60 = 40^\circ$
You are Right
$180 – 80 – 60 = 40$
$180 – 60 – 40 = 80$
$\angle$CHF = $180 – 80 – 60 = 40^\circ$
In the figure below, PQRS is a rectangle and QTUR is a square. PQT and SRU are straight line. Find $\angle$ SQU.
Sorry. Please check the correct answer below.
You are Right
The figure below shown four identical quadrants inside a square, ABCD. Find the perimeter of the shaded region. $ (Take \pi= \frac{22}{7})$
Sorry. Please check the correct answer below.
You are Right
In the figure below, AB and PQ are straight lines. Given that $\angle$BOG= 72 and$\angle$AOQ =108, find$\angle$POG.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
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 $\rightarrow$ $90^\circ - 58^\circ = 22^\circ$
The figure is made up of 1 square and 2 isosceles triangle. Find$\angle$ b.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$\angle$d + $\angle$e = $180^\circ = 79^\circ = 101^\circ$
$\angle$d + $\angle$e + $\angle$f + $\angle$g = $101^\circ + 202^\circ = 303^\circ$
You are Right
$\angle$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.
Sorry. Please check the correct answer below.
You are Right
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?
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$\frac{180 - 36}{1 + 2} = 48$ , $48 \times 2 = 96$
You are Right
$\frac{180 - 36}{1 + 2} = 48$ , $48 \times 2 = 96$
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.
Sorry. Please check the correct answer below.
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 $100^\circ + 55^\circ = 155^\circ$
You are Right
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, ST, UV, WX and YZ are straight lines. Which of the following angles, when added up, have the same value as $\angle$UOZ?
Sorry. Please check the correct answer below.
You are Right
The figure below shown an equilateral triangle XYZ and a straight line AB. Find$\angle$p
Sorry. Please check the correct answer below.
You are Right
Which one of the following statements is true?
Sorry. Please check the correct answer below.
You are Right
In the figure below not, drawn to scale, ABDE is a parallelogram. $\angle$ ACB =70$^\circ$. and $\angle$ BAC = 60 $^\circ$. Find $\angle$ EDC.
Sorry. Please check the correct answer below.
You are Right
The figure below is not drawn to scale. ABCD is a parallelogram and ABE is an isosceles triangle. Find $\angle$p.
Sorry. Please check the correct answer below.
10$^\circ$ $\rightarrow$ $180^\circ - 46^\circ - 46^\circ = 88^\circ$
$\angle$ $98^\circ - 88^\circ = 10^\circ$
You are Right
10$^\circ$ $\rightarrow$ $180^\circ - 46^\circ - 46^\circ = 88^\circ$
$\angle$ $98^\circ - 88^\circ = 10^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(a) 67$^\circ$ $\rightarrow$ $180 – 65 – 48 = 76^\circ$
(b) 132$^\circ$ $\rightarrow$ $\angle$ $KFE = 180 – 48 = 132^\circ$
(b) 132$^\circ$
You are Right
(a) 67$^\circ$ $\rightarrow$ $180 – 65 – 48 = 76^\circ$
(b) 132$^\circ$ $\rightarrow$ $\angle$ $KFE = 180 – 48 = 132^\circ$
In the figure below, not drawn to scale, WXYZ is a trapezium. AY, GW and YC are straight line. Find $\angle$YWX
Sorry. Please check the correct answer below.
$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$
You are Right
$180^\circ - 78^\circ = 102^\circ$
$180^\circ - 102^\circ – 42^\circ= 36^\circ$
$(180^\circ - 36^\circ) ÷ 2= 72^\circ$
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.
Sorry. Please check the correct answer below.
You are Right
The figure shown below is not drawn to scale. AB // CD and AD // BC. Find $\angle$m.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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$^\circ$
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.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
You are Right
$180 – 130 = 50$
$180 – 155 – 25$
$180 – 130 – 25 = 25$
$(180 - 50) ÷ 2 = 65^\circ$
In the figure below. ABCD is a rhombus, $\angle$ABC =102$^\circ $ and$\angle$CAE=19$^\circ $. Find $\angle$EAB.
Sorry. Please check the correct answer below.
20$^\circ$ $\rightarrow 180^\circ – 120^\circ = 78^\circ$
$78^\circ ÷ 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
You are Right
20$^\circ$ $\rightarrow 180^\circ – 120^\circ = 78^\circ$
$78^\circ ÷ 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
ABC and OBC are isosceles triangles. AB= AC, OB= OC, $\angle$ ABO=20$^\circ$ and $\angle$ BOC=80$^\circ$. Find$\angle$a.
Sorry. Please check the correct answer below.
$(180 – 80) ÷ 2 = 50$
50 + 20 = 70
$70 \times 2 = 140$
180 – 140 = 40$^\circ$
You are Right
$(180 – 80) ÷ 2 = 50$
50 + 20 = 70
$70 \times 2 = 140$
180 – 140 = 40$^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(a) $\angle$ $DAE = 90^\circ - 54^\circ = 36^\circ$
(b) 63$^\circ$ $\rightarrow$ $\angle$ $ACE = 180^\circ – (27^\circ+90^\circ) = 63^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(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 $\rightarrow$ $180^\circ - 40^\circ = 140^\circ$
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.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
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 shows an equilateral triangular piece of paper folded along line CX. $\angle$ACB is 22 $^\circ$. Find $\angle$ BCX.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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$^\circ$
In the figure below, AB and BC are two sides of a parallelogram ABCD. Measure and with down the value of $\angle$ABC.
Sorry. Please check the correct answer below.
You are Right
In the figure, ABCD is a rectangle and CEFG is a rhombus$\angle$EFG = 100. $^\circ$. and $\angle$DCG = 135 $^\circ$. Find$\angle$ BCE.
Sorry. Please check the correct answer below.
You are Right
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
Sorry. Please check the correct answer below.
You are Right
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
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$360 – 150 – 115 = 95$
$360 – 95 – 95 = 170$
$170 ÷ 2 = 85$
$\angle$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, AOC is a straight line $\angle$AOB = 159$^\circ$ and $\angle$COD = 63$^\circ$. What is the sum of$\angle$ AOD and$\angle$BOC?
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
You are Right
In the figure below, a rectangular piece of paper was folded as shown. Find $\angle$ DEB.
Sorry. Please check the correct answer below.
You are Right
The figure below consists of 3 straight line. What is the value of $\angle$a + $\angle$b +$\angle$c?
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
In the figure, ABCD is a rectangle. $\angle$AFE is 43$^\circ$. Find $\angle$FCD.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
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$^\circ$ $\rightarrow$ $\angle$ $EOB = 90^\circ 39^\circ = 51^\circ$
$\angle$ $AOC = 180^\circ - 39^\circ = 141^\circ$
$\angle$ $AOD = 360^\circ (180^\circ+50^\circ+51^\circ) = 360^\circ - 281^\circ = 79^\circ$
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.
Sorry. Please check the correct answer below.
You are Right
In the figure below, Find$\angle$X.
Sorry. Please check the correct answer below.
You are Right
In the figure below, not drawn to scale, LPRS is a parallelogram. ML = NL and NQ and MS are straight line. Find $\angle$PQR.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
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 figure below, ABCD is trapezium and BDE is an isosceles triangle, with DB = EB. Find $\angle$CDE.
Sorry. Please check the correct answer below.
$180^\circ - 50^\circ - 50^\circ = 80^\circ$
$110^\circ - 80^\circ = 30^\circ $
$50^\circ - 30^\circ = 20^\circ$
You are Right
$180^\circ - 50^\circ - 50^\circ = 80^\circ$
$110^\circ - 80^\circ = 30^\circ $
$50^\circ - 30^\circ = 20^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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 $\rightarrow$ $180^\circ - 93^\circ - 74^\circ = 13^\circ$
In the figure, CD and MN are straight lines. Which two angles add up to 90$^\circ$ ?
Sorry. Please check the correct answer below.
You are Right
In the figure, AB, CD, EF and GH are straight line and EF is parallel to GH. Which of the following statements is correct?
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
You are Right
In the figure below, a rectangular piece of paper is folded at a corner as shown Find $\angle$m.
Sorry. Please check the correct answer below.
You are Right
The figure is not drawn to scale. ABCD is a trapezium. Find $\angle$ADC.
Sorry. Please check the correct answer below.
You are Right
In the diagram below, ABCD is a square and BCD is an equilateral triangle. CFE is a straight lin. Find $\angle$AEF.
Sorry. Please check the correct answer below.
You are Right
A rectangular piece of paper was folded as shown below. Find $\angle$a.
Sorry. Please check the correct answer below.
You are Right
In the figure, find the sum of $\angle$a, $\angle$b, $\angle$c and $\angle$d.
Sorry. Please check the correct answer below.
$180 – 65 = 115$
$180 ÷ 3 \times 2 = 120$
$120 + 115 = 235$
You are Right
$180 – 65 = 115$
$180 ÷ 3 \times 2 = 120$
$120 + 115 = 235$
In the diagram below, ABCD is a trapezium and BCE is an equilateral triangle. AB // GF is a straight line. Find $\angle$ABC.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$\angle$CBD $\rightarrow$ $180^\circ - 54^\circ - 86^\circ = 40^\circ$
AC = CB = diameter
$\angle$BAC = $\angle$ABC = 58$^\circ$
$\angle$ABD $\rightarrow$ $58^\circ – 40^\circ = 18^\circ$
You are Right
$\angle$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 diagram below, EF is parallel to GH and JF and KE are straight line. Find$\angle$KGH.
Sorry. Please check the correct answer below.
$180 – 145 = 35$ $\angle$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$^\circ$
$180 – 35 – 68 = 77^\circ$
The figure below shows a cube. Give that A, B and c are points at the corners of the cube, what is $\angle$ ABC?
Sorry. Please check the correct answer below.
AC = AB = BC
Triangle ABC is an Equilateral Triangle, $\angle$ ABC = 60$^\circ$
You are Right
AC = AB = BC
Triangle ABC is an Equilateral Triangle, $\angle$ ABC = 60$^\circ$
The figure below is made up of a triangle and a trapezium. Find the sum of all the unknown marked angle in the figure.
Sorry. Please check the correct answer below.
You are Right
In the figure below, not drawn to scale, JKLM is a parallelogram. MKN is a straight line find the value of $\angle$LKN.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
$55^\circ - 30^\circ = 25^\circ$
$180^\circ - 55^\circ - 55^\circ = 70^\circ$
$180^\circ - 70^\circ - 30^\circ = 80^\circ$
You are Right
$55^\circ - 30^\circ = 25^\circ$
$180^\circ - 55^\circ - 55^\circ = 70^\circ$
$180^\circ - 70^\circ - 30^\circ = 80^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
$a 90^\circ - 75^\circ = 15^\circ$
$\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 two identical triangle. TSR and QSR. $\angle$QST = 148$^\circ$. Find $\angle$QSR.
Sorry. Please check the correct answer below.
106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
You are Right
106$^\circ$ $\rightarrow$ $(360 -148) ÷ 2 = 106^\circ$
In the figure, CDE and CDF are isosceles triangle. $\angle$DCF is three times as large as $\angle$EDF. Find $\angle$DFE.
Sorry. Please check the correct answer below.
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$^\circ$ $\rightarrow$ $4U + 3U + 3U = 10U$
$1U = 180^\circ ÷ 10 = 18^\circ$
$3U = 18^\circ \times 3 = 54^\circ$
$1U = 18^\circ$
$\angle$ $DEF = 180^\circ – 18^\circ = 54^\circ = 108^\circ$
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.
Sorry. Please check the correct answer below.
$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
You are Right
$(180^\circ - 90^\circ - 73^\circ) \times 2 = 34^\circ$
$45^\circ + (180^\circ - 34^\circ - 90^\circ) = 101^\circ$
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.
Sorry. Please check the correct answer below.
$\angle$ 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 below is not drawn to scale. ABCD is a trapezium and $\angle$ BDC = $\angle$ ADB. Find $\angle$ CAD.
Sorry. Please check the correct answer below.
120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$
You are Right
120$^\circ$ $\rightarrow$ $360 – 280 = 80$
$80 ÷ 2 = 40$
$\angle$ CAD = $180^\circ - 40^\circ - 20^\circ = 120^\circ$
AOB is a straight line as shown in the figure. $\angle$a = $\angle$b Find $\angle$a
Sorry. Please check the correct answer below.
32$^\circ$ $\rightarrow$ $64÷ 2 = 32$
You are Right
32$^\circ$ $\rightarrow$ $64÷ 2 = 32$
In the figure below, XYZ is an isosceles triangle. XY = YZ and AB// XY. AWY and AVB are straight line. Find $\angle$ t.
Sorry. Please check the correct answer below.
$\angle$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, 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$^\circ$
Sorry. Please check the correct answer below.
(a) $180 – 60 – 90 = 30$
$30 ÷ 2 = 15$
$45 – 15 = 30$
(b) $60 + 45 = 105^\circ$
(b) 105$^\circ$
You are Right
(a) $180 – 60 – 90 = 30$
$30 ÷ 2 = 15$
$45 – 15 = 30$
(b) $60 + 45 = 105^\circ$
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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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$^\circ$
In the diagram below, ABC is an equilateral triangle. BD and EF are straight line. Find $\angle$y
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
You are Right
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.
Sorry. Please check the correct answer below.
You are Right
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$^\circ$
Sorry. Please check the correct answer below.
(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 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$^\circ$
Sorry. Please check the correct answer below.
(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$^\circ$
You are Right
(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 = $180^\circ - 85^\circ = 95^\circ$
In the figure not scale, SU = SQ, $\angle$ TQP = 34$^\circ$, $\angle$ SQR = 91$^\circ$, and PRQ is parallel to TS. Find $\angle$ TSU.
Sorry. Please check the correct answer below.
You are Right
PSLE-Math.com
Start winning at math! Be part of Singapore PSLE Math community and Unleash the Your Maths Potential!
General
2025 © All Rights Reserved. Privacy Policy | Terms of Service
What subject area do you need help in?
Select level
Select subject
