# Question #4395c

Aug 20, 2016

$\frac{h \cos \left(\alpha - \beta\right) \sin \alpha}{\sin} \beta$

#### Explanation:

Given

• $H \to \text{Height of Tower} = C D$

• $h \to \text{Height of Pole} = A B$

• $\angle C B D \to \text{Angle of elevation top of Tower from B } = \alpha$

• $\angle A C B \to \text{Angle subtended by the pole at C } = \beta$

• $\angle C A E = \left(\alpha - \beta\right)$

• $\text{Let } A E = B D = b$

$\text{Now for } \Delta C B D , \frac{C D}{B D} = \frac{H}{b} = \tan \alpha \ldots \ldots \left(1\right)$

$\text{And for } \Delta C A E , \frac{C E}{A E} = \frac{H - h}{b} = \tan \left(\alpha - \beta\right) \ldots \ldots \left(2\right)$

Dividing (2) by (1) we get

$\frac{H - h}{H} = \tan \frac{\alpha - \beta}{\tan} \alpha$

$\implies 1 - \frac{h}{H} = \frac{\sin \left(\alpha - \beta\right) \cos \alpha}{\cos \left(\alpha - \beta\right) \sin \alpha}$

$\implies \frac{h}{H} = 1 - \frac{\sin \left(\alpha - \beta\right) \cos \alpha}{\cos \left(\alpha - \beta\right) \sin \alpha}$

$\implies \frac{h}{H} = \frac{\cos \left(\alpha - \beta\right) \sin \alpha - \sin \left(\alpha - \beta\right) \cos \alpha}{\cos \left(\alpha - \beta\right) \sin \alpha}$

$\implies \frac{h}{H} = \sin \frac{\alpha - \alpha + \beta}{\cos \left(\alpha - \beta\right) \sin \alpha}$

$\implies \frac{h}{H} = \sin \frac{\beta}{\cos \left(\alpha - \beta\right) \sin \alpha}$

$\implies \frac{H}{h} = \frac{\cos \left(\alpha - \beta\right) \sin \alpha}{\sin} \beta$

$\implies H = \frac{h \cos \left(\alpha - \beta\right) \sin \alpha}{\sin} \beta$