Question

how to prove that :
$\frac{d}{ \cot( \alpha ) + \cot(θ) } = h$
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Answer 1

Answer:

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Answer 2

The segment of length $h$ is perpendicular to the base.

Take the length of the divided base as $a$ and $b$.

$\implies h=a\tan\theta$

$\implies h=b\tan\alpha$

This gives,

$\implies a=h\cot\theta$

$\implies b=h\cot\alpha$

However, the length of the base is $d=a+b$.

(adding the two equations)

$\implies a+b=h(\cot\theta+\cot\alpha)$

(as $d=a+b$)

$\implies d=h(\cot\theta+\cot\alpha)$

(dividing both sides by $\cot\alpha+\cot\theta$)

$\implies h=\dfrac{d}{\cot\alpha+\cot\theta}$

Hence, proved.