In a triangle ABC, angle B = (π/3) and angle C = (π/4). Let D divides BC internally in the ratio 1: 3, then ( sin angle BAD / sin angle CAD) is equal to

Question

In a triangle ABC, angle B = (π/3) and angle C = (π/4). Let D divides BC internally in the ratio 1: 3, then ( sin angle BAD / sin angle CAD) is equal to (A)  (1/ √ 6) (B)  (1/ √ 3) (C) (1/3) (D) √ (2/3)

Tardigrade Admin · Accepted Answer

In triangle ABD, applying sine rule, we get = (AD/ sin π / 3) = ( x/sin α) ⇒ AD = (√ 3 /2) x sin α ...(i) and in triangle ACD, applying sine rule, we get (AD/ sin π / 4) = ( 3x/sin β) ⇒ AD = ( 3/ √ 2) x sin β ...(i) From Eqs. (i) and (ii). ( √ 3 x/ 2 sin α) = ( 3x/ √ 2 sin β) ⇒ ( sin α / sin β) = (1/ √ 6)

Q. In a $△ A BC$ , $∠ B = \frac{π}{3} an d ∠ C = \frac{π}{4} .$ Let D divides BC
internally in the ratio 1: 3, then $\frac{s in ∠ B A D}{s in ∠ C A D}$ is equal to

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

Q. In a △ABC, ∠B=3π​and∠C=4π​. Let D divides BC internally in the ratio 1: 3, then sin∠CADsin∠BAD​ is equal to

6​1​

3​1​

31​

32​​

In △ ABD, applying sine rule, we get = sinπ/3AD​=sinαx​ ⇒AD=23​​xsinα ...(i) and in △ ACD, applying sine rule, we get sinπ/4AD​=sinβ3x​ ⇒AD=2​3​xsinβ ...(i) From Eqs. (i) and (ii). 2sinα3​x​=2​sinβ3x​ ⇒sinβsinα​=6​1​

Q. In a $△ A BC$ , $∠ B = \frac{π}{3} an d ∠ C = \frac{π}{4} .$ Let D divides BC
internally in the ratio 1: 3, then $\frac{s in ∠ B A D}{s in ∠ C A D}$ is equal to

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{3}$

$\frac{2}{3}$