The two adjacent sides of a cyclic quadrilate | vedclass

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(C) Let the sides be $2, 5, a, b$ in order. The angle between sides $2$ and $5$ is $60^{\circ}$. Let $c$ be the diagonal opposite to the $60^{\circ}$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Let the sides be $2, 5, a, b$ in order. The angle between sides $2$ and $5$ is $60^{\circ}$. Let $c$ be the diagonal opposite to the $60^{\circ}$ angle.Using the Law of Cosines in the first triangle:$c^2 = 2^2 + 5^2 - 2(2)(5)\cos(60^{\circ}) = 4 + 25 - 20(0.5) = 29 - 10 = 19$.So,$c = \sqrt{19}$.Since the quadrilateral is cyclic,the opposite angle to $60^{\circ}$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.Using the Law of Cosines in the second triangle with sides $a, b$ and diagonal $c$:$c^2 = a^2 + b^2 - 2ab\cos(120^{\circ}) \implies 19 = a^2 + b^2 - 2ab(-0.5) \implies a^2 + b^2 + ab = 19$.The area of the quadrilateral is the sum of the areas of the two triangles:$	ext{Area} = \frac{1}{2}(2)(5)\sin(60^{\circ}) + \frac{1}{2}ab\sin(120^{\circ}) = 4\sqrt{3}$.$5\left(\frac{\sqrt{3}}{2}ight) + \frac{ab}{2}\left(\frac{\sqrt{3}}{2}ight) = 4\sqrt{3}$.Dividing by $\frac{\sqrt{3}}{2}$: $5 + \frac{ab}{2} = 8 \implies \frac{ab}{2} = 3 \implies ab = 6$.Now,$a^2 + b^2 = 19 - ab = 19 - 6 = 13$.We have $a^2 + b^2 = 13$ and $ab = 6$. Solving these,we find $a=2, b=3$ (or vice versa).The perimeter is $2 + 5 + a + b = 7 + 2 + 3 = 12$.

List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the area of the quadrilateral is $4\sqrt{3}$,then the perimeter of the quadrilateral is

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