Solveeit Logo
Login

Question

Question: In a triangle ABC, \(\angle B = \frac { \pi } { 3 }\) and \(\angle C = \frac { \pi } { 4 }\) andD d...

In a triangle ABC, ∠B=π3\angle B = \frac { \pi } { 3 } and ∠C=π4\angle C = \frac { \pi } { 4 } andD divides BC

internally in the ratio 1 : 3. Then sin⁡∠BADsin⁡∠CAD\frac { \sin \angle B A D } { \sin \angle C A D } is equal to

A

13\frac { 1 } { 3 }

B

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

C

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

D

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

Answer

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

Explanation

Solution

Let ∠BAD=α,∠CAD=β\angle B A D = \alpha , \angle C A D = \beta

In △ADB\triangle A D B, applying sine formulae, we get

xsin⁡α=ADsin⁡(π3)\frac { x } { \sin \alpha } = \frac { A D } { \sin \left( \frac { \pi } { 3 } \right) } ........(i)

In △ADC\triangle A D C applying sine formulae, we get,

3xsin⁡β=ADsin⁡(π4)\frac { 3 x } { \sin \beta } = \frac { A D } { \sin \left( \frac { \pi } { 4 } \right) } ..........(ii)

Dividing (i) by (ii), we get,

⇒ xsin⁡α×sin⁡β3x=ADsin⁡(π3)×sin⁡(π4)AD\frac { x } { \sin \alpha } \times \frac { \sin \beta } { 3 x } = \frac { A D } { \sin \left( \frac { \pi } { 3 } \right) } \times \frac { \sin \left( \frac { \pi } { 4 } \right) } { A D } ⇒ sin⁡β3sin⁡α=1232=23\frac { \sin \beta } { 3 \sin \alpha } = \frac { \frac { 1 } { \sqrt { 2 } } } { \frac { \sqrt { 3 } } { 2 } } = \sqrt { \frac { 2 } { 3 } }

⇒ sin⁡βsin⁡α=323=6\frac { \sin \beta } { \sin \alpha } = 3 \sqrt { \frac { 2 } { 3 } } = \sqrt { 6 }

∴\therefore sin⁡∠BADsin⁡∠CAD=sin⁡αsin⁡β=16\frac { \sin \angle B A D } { \sin \angle C A D } = \frac { \sin \alpha } { \sin \beta } = \frac { 1 } { \sqrt { 6 } }