In triangle ABC,if a=2(sqrt{3}+1),B=45^{circ} | vedclass

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

Question

(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B}

Vedclass · Accepted Answer

$6+2 \sqrt{3}$

Vedclass · Answer

(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$a = 2(\sqrt{3}+1)$,$\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.So,$\frac{2(\sqrt{3}+1)}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = 4\sqrt{2}$.Thus,$b = 4\sqrt{2} \sin 45^{\circ} = 4\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 4$.$c = 4\sqrt{2} \sin 60^{\circ} = 4\sqrt{2} 	imes \frac{\sqrt{3}}{2} = 2\sqrt{6}$.Area $= \frac{1}{2} ab \sin C = \frac{1}{2} 	imes 2(\sqrt{3}+1) 	imes 4 	imes \sin 60^{\circ} = 4(\sqrt{3}+1) 	imes \frac{\sqrt{3}}{2} = 2(3+\sqrt{3}) = 6+2\sqrt{3}$.

In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

Similar Questions

In $\triangle ABC$, with usual notations, if $a = 4$, $b = 5$, and $c = 6$, then $\cos A$ and $\cos C$ are calculated using the Law of Cosines. Find the value of $\cos C$ and $\cos A$ to determine the relationship between $\angle C$ and $\angle A$.

In $\Delta ABC$, if $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$, then

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\frac{\cos A + 3 \cos C}{\cos B} = $

With usual notations,if in $\triangle ABC$,$s$ is the semi-perimeter and $(s-a)(s-b)=s(s-c)$,then $\triangle ABC$ is

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

Vedclass Test Series

Exam Paper Generator

Online Exam Module