Angles of Elevation and Depression, Short questions (Question 4 & 5)


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 15o.
The angle of depression of M from N is 35o.
Calculate the distance, in m, from K to L.
 
Solution:
tan A P N = A N P A tan 15 o = A N 4 A N = 4 × 0.268 A N = 1.072 m Length of B N = 2 + 1.072 = 3.072 c m tan B M N = B N B M tan 35 o = 3.072 B M B M = 3.072 0.700 = 4.389 Distance of K L = 4.389 m


Question 5:
Diagram below shows two vertical poles JM and LN, on a horizontal plane.
The angle of elevation of M from K is 70oand the angle of depression of K from N is 40o.
Find the difference in distance, in m, between JK and KL.
 
Solution:
tan J K M = 14 J K J K = 14 tan 70 o J K = 5.096 m tan L K N = 8 K L K L = 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

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