the angle of elevation of the …

 
Let height of tower = h m
Let distance {tex}BC = x \ m{/tex}
In {tex}\Delta ABC{/tex}
{tex}\tan {63^o} = \frac{{AB}}{{BC}}{/tex}
{tex} \Rightarrow 1.9626 = \frac{h}{x}{/tex}
{tex} \Rightarrow x = \frac{h}{{1.9626}}{/tex}
{tex} \Rightarrow x = 0.5095h{/tex} ...(i)
In {tex}\Delta ABD{/tex}
{tex}tan32^\circ= \frac{h}{{100 + x}}{/tex}
{tex}\Rightarrow 0.6248= \frac{h}{{100 + x}}{/tex}
{tex} \Rightarrow h = 62.48 + 0.6248x{/tex}
{tex} \Rightarrow h = 62.48 + 0.6248 \times 0.5059h{/tex}
{tex} \Rightarrow 0.6817h = 62.48{/tex}
{tex} \Rightarrow h = \frac{{62.48}}{{0.6817}} = 91.65m{/tex}
Put value of h in Eq(i) , we get 
{tex}x = 0.5095 \times 91.65 = 46.69m{/tex}
{tex}\therefore{/tex} Height of tower = 91.65 m
Distance of first position from tower {tex}= 100 + x{/tex}
{tex}= 100 + 46.69 \\ = 146.69 m{/tex}