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:
tanAPN= AN PA tan 15 o = AN 4 AN=4×0.268 AN=1.072m Length of BN=2+1.072=3.072cm tanBMN= BN BM tan 35 o = 3.072 BM BM= 3.072 0.700 =4.389 Distance of KL=4.389m

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

Solution:
tanJKM= 14 JK JK= 14 tan 70 o JK=5.096m tanLKN= 8 KL KL= 8 tan 40 o KL=9.534m Difference in distance of JK and KL =9.5345.096 =4.438m