In a ABC- - B-3- C -4 and D divides BC internally in the ratio 1-3 then sin- BADsin- CAD is equal to-

The correct option is C 1√(6) Consider #916;ABD and #916;ACD: P= #916;ABD #916;ACD = 1 2 AB #8901;ADsin #8736;BAD 1 2 AC ...

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Standard VII

Mathematics

Angle Sum Property

In a ABC, ∠...

Question

In a $△ A B C, ∠ B = \frac{π}{3}, ∠ C = \frac{π}{4}$ and $D$ divides $B C$ internally in the ratio $1 : 3$ then $\frac{sin ∠ B A D}{sin ∠ C A D}$ is equal to:

\frac{1}{\sqrt{2}}

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\frac{1}{\sqrt{3}}

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\frac{1}{\sqrt{6}}

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\sqrt{6}

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Solution

The correct option is C $\frac{1}{\sqrt{6}}$

$C o n s i d e r Δ A B D a n d Δ A C D :$

$P = \frac{Δ A B D}{Δ A C D} = \frac{\frac{1}{2} A B \cdot A D s i n ∠ B A D}{\frac{1}{2} A C \cdot A D s i n ∠ C A D}$

$P = \frac{A B s i n ∠ B A D}{A C s i n ∠ C A D} ⟹ (1)$

$U s i n g t h e L a w o f S i n e s,$

$\frac{A B}{s i n 45 °} = \frac{A C}{s i n 60 °}$

$\frac{A B}{A C} = \frac{s i n 45 °}{s i n 60 °} = \frac{\sqrt{2}}{\sqrt{3}} ⟹ (2)$

$F r o m (1) a n d (2) :$

$P = \frac{\sqrt{2} s i n B A D}{\sqrt{3} s i n C A D} ⟹ (3)$

$S i n c e t h e s e t r i a n g l e s h a v e t h e s a m e a l t i t u d e,$

$P = \frac{B D}{D C} = \frac{1}{3} ⟹ (4)$

$E q u a t i n g (3) a n d (4) :$

$\frac{s i n ∠ B A D}{s i n ∠ C A D} = \frac{1}{\sqrt{6}}$

Suggest Corrections

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In a △ABC,∠B=π3,∠C=π4 and D divides BC internally in the ratio 1:3 then sin∠BADsin∠CAD is equal to:

In a $△ A B C, ∠ B = \frac{π}{3}, ∠ C = \frac{π}{4}$ and $D$ divides $B C$ internally in the ratio $1 : 3$ then $\frac{sin ∠ B A D}{sin ∠ C A D}$ is equal to: