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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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The figure below is not drawn to scale. ABCD is a parallelogram and ABE is an isosceles triangle. Find $\angle$p.
10$^\circ$ $\rightarrow$ $180^\circ - 46^\circ - 46^\circ = 88^\circ$ $\angle$ $98^\circ - 88^\circ = 10^\circ$
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?
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
(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 diagram below, ABCD is a trapezium and AEF is an equilateral triangle. Find (a) $\angle$BAD. (b) $\angle$DFE.
(a) 80$^\circ$ (b) $180^\circ – 80^\circ - 60^\circ = 40^\circ$ $60^\circ – 40 = 20^\circ$
In the figure below, not drawn to scale, LPRS is a parallelogram. ML = NL and NQ and MS are straight line. Find $\angle$PQR.
In the figure below, AB and PQ are straight lines. Given that $\angle$BOG= 72 and$\angle$AOQ =108, find$\angle$POG.
In the figure below. ABFE is a parallelogram and BCDE is trapezium. Given that $\angle$AFE =75 $^\circ$, Find $\angle$y.
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.
In the figure below, a rectangular piece of paper was folded as shown. Find $\angle$ DEB.
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