Question

1.The angle of depression of an object from the top of a hill is 600,If the height of the hill is 250m,how far is the object from the foot of the hill ? 2.The angle of elevation of point P from point Q is 400. PQ=45 km.How high is point P above the level of point Q ?

Answer 1

1. if the angle of depression is 60° the angle b/n the line of sight an d the hill will be 30°. and lets say the horizontal distance b/n the obj n the hill is x.

then tan 30= x/250m

x = 0.577×250m = 144.34 m

2. sin 40= opp/hyp = h/45km

h= 0.64×45km = 28.8km