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WAEC Mathematics 2016 Paper

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Question : 27

Total: 49

In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48^o and < TXW = 60^o.Find < TQU.

120^o
108^o
72^o
60^o

Solution:

Concept:

Use the angle properties of parallel lines and the angle sum of a triangle.

Explanation:

Given

\mathrm{PR} \parallel \mathrm{SV} \parallel \mathrm{WY}

and

\mathrm{TX} \parallel \mathrm{QY}

.

Since

\angle TXW = 60^\circ

, the corresponding angle is

\angle TUQ = 60^\circ

.

Also,

\angle PQT = 48^\circ

.

Because

\mathrm{PR} \parallel \mathrm{WY}

and

\mathrm{QT}

is a transversal,

\angle QTU = 48^\circ

(alternate angles).

In

\triangle TQU

, the sum of angles is

180^\circ

.

So,

\angle TQU + 60^\circ + 48^\circ = 180^\circ

.

Therefore,

\angle TQU = 180^\circ - 108^\circ = 72^\circ

.

Answer:

C.

72^\circ

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