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JAMB Mathematics 2020 Paper

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Question : 17

Total: 40

In this fiqure, PQ = PR = PS and SRT = 68 $^o$ . Find QPS

136 $^o$
124 $^o$
112 $^o$
68 $^o$

Solution:

Since PQRS is quadrilateral

2y + 2x + QPS = 360 $^o$

i.e. (y + x) + QPS = 360 $^o$

QPS = 360 $^o$ - 2 (y + x)

But x + y + 68 $^o$ = 180 $^o$

There; x + y = 180 $^o$ - 68 $^o$ = 112 $^o$

QPS = 360 - 2(112 $^o$ )

= 360 $^o$ - 224 = 136 $^o$

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