When the sun s altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?
Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, $\angle$ABD = 30°, $\angle$ACD = 60°,
Let CD = x, AD = h
From the right $\triangle$ CDA,
tan60°=$\dfrac{AD}{CD}$
√3=$\dfrac{h}{x} $ ⋯(eq:1)
From the right $\triangle$ BDA,
tan30°=$\dfrac{AD}{BD}$
$\dfrac{1}{\sqrt{3}}=\dfrac{h}{70+x} $ ⋯(eq:2)
$\dfrac{eq:1}{eq:2}⇒\dfrac{\sqrt{3}}{\left( \dfrac{1}{\sqrt{3}} \right)}=\dfrac{\left( \dfrac{h}{x}\right)}{\left( \dfrac{h}{70+x}\right)}$
⇒3=$\dfrac{70+x}{x}$
⇒2x=70
⇒x=35
Substituting this value of x in eq:1, we have
√3=$\dfrac{h}{35}$
⇒h=35√3=35×1.73
=60.55≈60.6