SSC CGL Tier II 21 Sep 2014 Paper

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Question : 42 of 100

Marks: +1, -0

In a ΔABC, ∠B = p 3 , ∠C =

\π/ 4

and D divides BC internally in the ratio 1 : 3 then

{\sin∠ B A D}/{\sin ∠ C A D}

is equal to

$1 /√2$
$1/√ 3$
$1/√ 6$
$√ 6$

Solution:

(3)

∠ B= {\π}/{3}, ∠= {\π}/{4}

and

{B D}/{D C}= {1}/{3}

From ΔABD,

{B D}/{\sin B A D}= {A D}/{\sin A B D}

⇒{B D}/{\sin B A D}= {A D}/{\sin {\π}/{3}}

⇒A D= {√{3}}/{2}.{B D}/{\sin BAD}...(i)

From ΔADC,

{ {CD}}/{\sin {DAC}}= { {AD}}/{\sin {ACD}}

⇒{ {CD}}/{\sin {DAC}}= { {AD}}/{\sin {\π}/{4}}

⇒A D= {1}/{√{2}} .{C D}/{\sin D A C} ...(ii)

From equations (i) and (ii),

{√{3}}/{2}.{ {BD}}/{\sin {BAD}}= {1}/{√{2}}.{ {CD}}/{\sin {DAC}}

⇒{\sin B A D}/{\sin D A C}= { {√{3}}/{2}}/{ {1}/{√{2}}} ×{B D}/{C D}

⇒{\sin B A D}/{\sin D A C}= {√{3}}/{2} ×√{2} ×{1}/{3}

= {1}/{√{2} ×√{3}}= {1}/{√{6}}

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