From the foot of a hill the angle of elevation of the tip of a tower is found to be 45°. After walking 2 km upwards along the slope of the hill ,which is inclined at 30°, the same is formed to be 60°. Find the height of the tower
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Let AB is the Tower of height = h = 50 m.
And, let the Height of Hill CD = H m. Distance between The root of the tower and hill = BC
Now,
In AABC
ZC = 30°
TAN(C) = AB/BC
⇒ TAN(30) = 50/BC
⇒1/√3 = 50/BC
⇒ BC= 50-√3 m.
Now, In ABCD,
ZB = 60°
Tan(B) = CD/BC
⇒ Tan(60) = H/BC
⇒ BC√3 = H
⇒ H=50-√3*√3 = 150 m.
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