Question

Solve this fast . I need full solution !​

Answer 1

Step-by-step explanation:

$refer \: to \: the \: attachment$

Answer 2

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CD = 100m

∆BDE = 30°

∆BCA = 45°

CA = x and EB = h

Consider ∆BDE

Tan 30° = h/x

$\frac{1}{ \sqrt{3} } = \frac{h}{x} => \: x = \sqrt{3} h........{1}$

Consider ∆BCA

Tan 45° = BA/CA

$1 = \frac{100 + h}{x} => \: x = 100 + h$

$\sqrt{3} h = 100 + h \: (from \: 1)$

$( \sqrt{3} - 1)h = 100 \: => h = \frac{100}{ \sqrt{3} - 1 } \times \frac{ \sqrt{3 + 1} }{ \sqrt{3 + 1} } = \frac{100 \sqrt{3 + 1} }{2} = 50( \sqrt{3} + 1)$

$height \: of \: a \: rock = 100 + h = 100 + 50 (\sqrt{3} + 1)$

$= 100 + 50 \sqrt{3} + 50$

$=( 105 + 50\sqrt{3} )m = 50 (3 + \sqrt{3})m$

$\huge { \underline { \mathtt { \red{hope} \: { \orange{it} \: { \pink{help}}}}}}$