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Gurvinder Kaur 7 years, 10 months ago
Let the height of the opposite house be DC=h metre
In r.t. {tex}\Delta ADE,{/tex}
{tex}\tan \,60^\circ = \frac{{DE}}{{AE}}{/tex}
{tex}\sqrt 3 = \frac{{h - 60^\circ }}{{AE}}{/tex}
{tex}AE = \frac{{h - 60}}{{\sqrt 3 }}\,..(i){/tex}
If r.t. {tex}\Delta ACE,{/tex}
{tex}\tan \,45^\circ = \frac{{CE}}{{AE}}{/tex}
{tex}AE = 60\,..(ii){/tex}
Comparing (i) and (ii), we get
{tex}\frac{{h - 60}}{{\sqrt 3 }} = 60{/tex}
{tex}h - 60 = 60\sqrt 3 {/tex}
{tex}h = 60\sqrt 3 + 60{/tex}
{tex}h = 60(1 + \sqrt 3 ){/tex}
Therefore, height of the opposite house is {tex}60(1 + \sqrt 3 ){/tex} metre.
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