The angle of elevation of a …
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Sia ? 6 years, 4 months ago
In {tex}\triangle{/tex}ADC,
{tex}\tan 30 ^ { \circ } = \frac { H - 200 } { x }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { H - 200 } { x }{/tex}
{tex}\Rightarrow \quad x = \sqrt { 3 } {/tex}(H - 200) m. ............(1)
In {tex}\triangle{/tex}ADF,
{tex}\tan 60 ^ { \circ } = \frac { H + 200 } { x }{/tex}
{tex}\sqrt { 3 } = \frac { H + 200 } { x }{/tex}
{tex}\sqrt3x=H+200{/tex}...............(2)
here H is height of cloud above the lake.
Substituting the value of x from (1) into (2) we get
3(H - 200) = H + 200
3H - 600 = H + 200
2H = 800
H = 400 m
So height of the cloud above the lake is 400 m.
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