If the angle of elevation of …
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Sia ? 6 years, 6 months ago
Let the height of tower be h.
From {tex}\triangle A B D{/tex} {tex}\frac { h } { a } = \tan 30 ^ { \circ }{/tex}
{tex}\therefore {/tex} {tex}h = a \times \frac { 1 } { \sqrt { 3 } } = \frac { a } { \sqrt { 3 } } \quad \ldots ( \mathrm { i } ){/tex}
From {tex}\triangle A C D,{/tex} {tex}\frac { h } { b } = \tan 60 ^ { \circ }{/tex}
{tex}h = b \times \sqrt { 3 } = b \sqrt { 3 } \quad \ldots ( \mathrm { ii } ){/tex}
From (i) {tex}a = \sqrt { 3 } h{/tex}
From (ii) {tex}b = \frac { h } { \sqrt { 3 } }{/tex}
{tex}\therefore \quad a \times b = \sqrt { 3 } h \times \frac { h } { \sqrt { 3 } }{/tex}
{tex}\Rightarrow \quad h ^ { 2 } = a b{/tex}
{tex}\Rightarrow \quad h = \sqrt { a b }{/tex}
Hence, the height of the tower = {tex}\sqrt { a b }{/tex}
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