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BECE Mathematics 2014 Paper

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Question : 34

Total: 40

Use the diagram below to answer the question below.

Find the angle marked

d

$38^{\circ}$
$40^{\circ}$
$48^{\circ}$
$88^{\circ}$

Solution:

Angle

PRQ=e

(Vertically opposite angles)

Sum of angles in a triangle is

180^{\circ}

\;50+38+PRQ=180

\;88+PRQ=180

\;PRQ=180-88

\;PRQ=92

But angle

PRQ=e=92

132+

angle

RTS=180

(Sum of angles on straight line is

180^{\circ}

)

Angle RTS =

180-132

Angle

RTS=48

\;e+d+\;\text{ Angle }\;RTS=180

Sum of angles in a triangle is 180°

\;e=92

\;92+d+48=180

\;d+140=180

\;d=180-140

\;d=40^{\circ}

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