The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Answer with explanationAnswer: Option B
AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°,
i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.
Now, let AB be h m and BC be x m.
According to the question, DB is 40 m longer than BC.
So, DB = (40 + x) m
Now, we have two right triangles ABC and ABD.
In Δ ABC,
tan 60° = AB / BC
or, √3 = h / x ….. (1)
In Δ ABD,
tan 30° = AB / BD
i.e., 1 / √3 = h / (40 + x ) ….. (2)
From (1), we have
h = x√3
Putting this value in (2), we get
(x√3)√3 = x + 40,
i.e., 3x = x + 40
i.e., x = 20