Question-47354

Question Number 47354 by behi83417@gmail.com last updated on 08/Nov/18

Commented by behi83417@gmail.com last updated on 08/Nov/18

A B C, i s a e q u e l a t e r a l t r i a n g l e w i t h

A B = 2, a n d : A C^{'} = 3, {B A}^{'} = 2 .

\dots \dots CD = ?

Commented by ajfour last updated on 09/Nov/18

Commented by ajfour last updated on 09/Nov/18

γ = 30 ° \Rightarrow ∠ C = 90 °

A^{'} C = \sqrt{4^{2} - 2^{2}} = 2 \sqrt{3} \dots (i)

F r o m △ B C D

\frac{\sin β}{2} = \frac{\sin γ}{B D} & F r o m △ {B D A}^{'}

\frac{\sin γ}{B D} = \frac{\sin (β - γ)}{A^{'} D}

\Rightarrow A^{'} D = \frac{3 \sin (β - γ)}{\sin β} \dots . (i i)

F r o m △ C C^{'} D

\frac{\sin α}{C D} = \frac{\cos β}{C D} = \frac{\sin β}{5}

\Rightarrow \tan β = \frac{5}{C D} \dots (i i i)

U s i n g (i), (i i) f o r

C D + A^{'} D = A^{'} C

\Rightarrow C D + \frac{3 \sin (β - γ)}{\sin β} = 2 \sqrt{3} \dots (I)

l e t C D = x

T h e n u s i n g (i i i) & (I)

x + 3 (\cos γ - \frac{\sin γ}{\tan β}) = 2 \sqrt{3}

\Rightarrow x + 3 (\frac{\sqrt{3}}{2} - \frac{x}{10}) = 2 \sqrt{3}

o r \frac{7 x}{10} = \frac{\sqrt{3}}{2}

\Rightarrow x = C D = \frac{5 \sqrt{3}}{7} .

Commented by behi83417@gmail.com last updated on 09/Nov/18

s i r A j f o u r! t h a n k y o u v e r y m u c h

f o r n i c e w o r k . g o d b l e s s y o u s i r .