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WAEC Mathematics 2012 Paper

Question : 34

Total: 49

< QPS = 30^o, < QRP = 90^o and |PS| = x, Find x

Solution:

Concept:

Apply trigonometry in the right-angled triangle formed by points

Q

P

and

R

Explanation:

Point

S

lies on

PR

, so the full length is

PR = PS + SR = x + 8

In the right-angled triangle

QPR

\angle QRP = 90^\circ

The angle at

P

\angle QPS = 30^\circ

Using the tangent ratio,

\tan 30^\circ = \frac{QR}{PR} = \frac{10}{x+8}

Substitute

\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.5773

x + 8 = \frac{10}{0.5773} \approx 17.32

Subtract 8 from both sides:

x = 17.32 - 8 = 9.32

Answer:

x = 9.32

Therefore, the correct option is C.

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