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WAEC Mathematics 2019 Paper

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Question : 44

Total: 50

In the diagram, PQ is parallel to RS, < QFG = 105

^o

^o

. Find the value of n

40 $^o$
55 $^o$
75 $^o$
130 $^o$

Solution:

Concept:

Use alternate interior angles and the angle sum on a straight line.

Explanation:

Since

PQ \parallel RS

and

EG

is a transversal, alternate interior angles are equal.

So,

\angle EGR = \angle FEG = 50^\circ

.

On line

PQ

,

\angle QFG

and

\angle EFG

form a linear pair.

Thus,

\angle EFG = 180^\circ - 105^\circ = 75^\circ

.

In

\triangle EFG

, the sum of angles is

180^\circ

.

So,

\angle FGE = 180^\circ - 50^\circ - 75^\circ = 55^\circ

.

At point

G

, angles on the straight line

RGS

add up to

180^\circ

.

Therefore,

n + 55^\circ + 50^\circ = 180^\circ

.

Hence,

n = 180^\circ - 105^\circ = 75^\circ

.

Answer:

n = 75^\circ

, which is Option C.

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