In a triangle ABC,if b=7, c=4sqrt{3} and A=fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\fr

Vedclass · Accepted Answer

$\frac{7\sqrt{3}}{\sqrt{13}}$

Vedclass · Answer

(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\frac{\pi}{6})$.$a^2 = 49 + 48 - 56\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 97 - 84 = 13$.So,$a = \sqrt{13}$.By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Thus,$\sin B = \frac{b \sin A}{a} = \frac{7 \sin(\pi/6)}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.And $\sin C = \frac{c \sin A}{a} = \frac{4\sqrt{3} \sin(\pi/6)}{\sqrt{13}} = \frac{2\sqrt{3}}{\sqrt{13}}$.Therefore,$a \sin B \sin C = \sqrt{13} 	imes \frac{7}{2\sqrt{13}} 	imes \frac{2\sqrt{3}}{\sqrt{13}} = \frac{7\sqrt{3}}{\sqrt{13}}$.

In a triangle $ABC$,if $b=7, c=4\sqrt{3}$ and $A=\frac{\pi}{6}$,then $a \sin B \sin C =$

Similar Questions

In a $\triangle ABC$,if $\angle A = 60^{\circ}$,then $\frac{b}{c+a} + \frac{c}{a+b} = $

In $\triangle ABC$,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} =$

In a $\Delta PQR$,if $3 \sin P + 4 \cos Q = 6$ and $4 \sin Q + 3 \cos P = 1$,then the angle $R$ is equal to:

In a triangle $ABC$, with usual notations, if $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ and $\lambda \tan^2 \frac{A}{2} = 455$, then $\lambda = $

Let $ABC$ be a triangle in which $AB=BC$. Let $X$ be a point on $AB$ such that $AX:XB=AB:AX$. If $AC=AX$,then the measure of $\angle ABC$ equals (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module