Question

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?

Answer 1

Let AD be the tower, BD be the initial shadow and CD be the final shadow.

Given that BC = 70 m, ABD = 30°, ACD = 60°,

Let CD = x, AD = h

From the right  CDA, 
@@\tan 60° = \dfrac{\text{AD}}{\text{CD}}\\ \sqrt{3} = \dfrac{\text{h}}{\text{x}}\quad \cdots(eq:1)@@

From the right  BDA, 
@@\tan 30° = \dfrac{\text{AD}}{\text{BD}}\\ \dfrac{1}{\sqrt{3}} = \dfrac{\text{h}}{70 + \text{x}}\quad \cdots(eq:2)@@

@@\dfrac{eq:1}{eq:2} \Rightarrow \dfrac{\sqrt{3}}{\left(\dfrac{1}{\sqrt{3}}\right)}=\dfrac{\left(\dfrac{\text{h}}{\text{x}}\right)}{\left(\dfrac{\text{h}}{70+\text{x}}\right)}\\ \Rightarrow 3 = \dfrac{70 + \text{x}}{\text{x}}\\ \Rightarrow 2x = 70 \\ \Rightarrow x = 35@@

Substituting this value of x in eq:1, we have
@@\sqrt{3} = \dfrac{\text{h}}{35}\\ \Rightarrow \text{h} = 35\sqrt{3} = 35 \times 1.73 \\= 60.55 \approx 60.6 @@


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