Ln a triangle ABC- - B-3- C-4 and D divides BC internally in the ratio 1-3- Then sin- BAD-sin- CAD is equal to-

The correct option is A 1/√(6) sin∠ BADx = sin Bl sin∠ CAD3x = sin Cl 3sin∠ BADsin∠ CAD = √(3)21√(2) sin∠ BADsin∠ CAD = 1√(6) ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

ln a triangle...

Question

$l n$ a triangle $A B C, ∠ B = \frac{π}{3}, ∠ C = \frac{π}{4}$ 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{6}}

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\frac{1}{3}

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

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\sqrt{\frac{2}{3}}

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Solution

The correct option is A $\frac{1}{\sqrt{6}}$
$\frac{sin ∠ B A D}{x} = \frac{sin B}{l}$

$\frac{sin ∠ C A D}{3 x} = \frac{sin C}{l}$

$\frac{3 sin ∠ B A D}{sin ∠ C A D} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}$
$\frac{sin ∠ B A D}{sin ∠ C A D} = \frac{1}{\sqrt{6}}$

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Similar questions

Q. In a triangle

A B C, ∠ = π / 3

and

∠ C = π / 4

. Let

D

divide

B C

internally in the ratio

1 : 3

. Then

\frac{sin ∠ B A D}{sin ∠ C A D}

equals.

Q. 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:

Q. If in a

△ A B C, ∠ B = \frac{π}{3}

and

∠ C = \frac{π}{4},

let

D

divide

B C

internally in the ratio

1 : 3

.Then

\frac{sin (∠ B A D)}{sin (∠ C A D)}

is equal to

Q. In a

Δ A B C

∠ B = \frac{π}{3}

and

∠ C = \frac{π}{4}

. If point D divides BC internally in the ratio

1 : 3

, then

\frac{sin ∠ B A D}{sin ∠ C A D}

is equal to

Q. In triangle

A B C,

∠ B = \frac{π}{3}

and

∠ C = \frac{π}{4} .

Let

D

divide

B C

internally in the ratio

1 : 3

and the angles

∠ B A D = θ,

∠ C A D = ϕ .

The value of

\frac{{sin}^{2} ϕ}{{sin}^{2} θ} =

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ln a triangle ABC, ∠B=π3,∠C=π4 and D divides BC internally in the ratio 1:3. Then sin∠BADsin∠CAD is equal to:

The correct option is A 1√6sin∠BADx=sinBlsin∠CAD3x=sinCl3sin∠BADsin∠CAD=√321√2sin∠BADsin∠CAD=1√6

$l n$ a triangle $A B C, ∠ B = \frac{π}{3}, ∠ C = \frac{π}{4}$ and $D$ divides $B C$ internally in the ratio 1:3. Then $\frac{sin ∠ B A D}{sin ∠ C A D}$ is equal to:

The correct option is A $\frac{1}{\sqrt{6}}$
$\frac{sin ∠ B A D}{x} = \frac{sin B}{l}$

$\frac{sin ∠ C A D}{3 x} = \frac{sin C}{l}$

$\frac{3 sin ∠ B A D}{sin ∠ C A D} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}$
$\frac{sin ∠ B A D}{sin ∠ C A D} = \frac{1}{\sqrt{6}}$