Question

Plz solve as fast as possible to mark u as brainliest. Plz stepwise method

Answer 1

$\boxed{\huge{\red{\mathfrak{Answer}}}}$

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AD = DP = 15 m

Let CP → x

AP = BD ,

In ∆ ACP,

$tan \: 30 ° \: = \frac{PC}{ AP} \\ \\ AP = x \sqrt{3}$

But AP = BD , BD = x√3

In ∆ BCD ,

$tan \: 60° = \frac{CD}{ BD} \\ \\ \sqrt{3} = \frac{x + 15}{BD} \\ \\ BD \sqrt{3} = x + 15$

But BD = x√3 ( Proved Above )

$3x \: = x + 15 \\ \\ 2x = 15 \\ \\ x = \frac{15}{2} \\ \\ x = 7.5 \: m$

$\huge{\mathtt{CD = PC + PD}}$

$\huge{\mathtt{CD = 7.5 + 15}}$

$\huge{\mathtt{CD = 22.5 m}}$

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