SPM Practice (Short Questions)

Question 1:

Diagram below shows two vertical poles on a horizontal plane. J, K, L, M and N are five points on the poles such that KL = MN.

Name the angle of elevation of point J from point M.

Solution:

Angle of elevation of point J from point M = ∠KMJ

Question 2:

Diagram below shows two vertical poles JM and KL, on a horizontal plane.

The angle of depression of vertex K from vertex J is 42^o.

Calculate the angle of elevation of vertex K from M.

Solution:

JN = 10 tan 42^o = 9.004 m

NM = 25 – 9.004 = 15.996 m

KL = NM = 15.996 m

$\begin{array}{l} \tan ∠ K M L = \frac{K L}{M L} \\ = \frac{15.996}{10} \\ = 1.5996 \\ ∠ K M L = \tan^{- 1} 1.5996 \\ = 57^{o} 59^{'} \end{array}$

Question 3:

Diagram below shows a vertical pole KMN on a horizontal plane. The angle of elevation of M from L is 20^o.

Calculate the height, in m, of the pole.

Solution:

$\begin{array}{l} \tan 20^{o} = \frac{K M}{K L} \\ 0.3640 = \frac{K M}{15} \\ K M = 5.4 m \\ Height of the pole = 9 + 5.4 = 14.4 m \end{array}$