SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 4:

Diagram below shows three vertical poles JP, KN and LM, on a horizontal plane.

The angle of elevation of N from P is 15^o.

The angle of depression of M from N is 35^o.

Calculate the distance, in m, from K to L.

Solution:

$\begin{array}{l} \tan ∠ A P N = \frac{A N}{P A} \\ \tan 15^{o} = \frac{A N}{4} \\ A N = 4 \times 0.268 \\ A N = 1.072 m \\ Length of B N = 2 + 1.072 = 3.072 c m \\ \tan ∠ B M N = \frac{B N}{B M} \\ \tan 35^{o} = \frac{3.072}{B M} \\ B M = \frac{3.072}{0.700} = 4.389 \\ Distance of K L = 4.389 m \end{array}$

Question 5:

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

The angle of elevation of M from K is 70^oand the angle of depression of K from N is 40^o.

Find the difference in distance, in m, between JK and KL.

Solution:

$\begin{array}{l} \tan ∠ J K M = \frac{14}{J K} \\ J K = \frac{14}{\tan 70^{o}} \\ J K = 5.096 m \\ \tan ∠ L K N = \frac{8}{K L} \\ K L = \frac{8}{\tan 40^{o}} \\ K L = 9.534 m \\ Difference in distance of J K and K L \\ = 9.534 - 5.096 \\ = 4.438 m \end{array}$