Answer: Option A. -> 1√5
:
A
We know, $sin\left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b.c}}$
Where"s" is the semiperimeter and b & c are the sides of a triangle.
We'll do the appropriate substitution -
$sin\left(\frac{A}{2}\right)=\sqrt{\frac{(6-3)(6-5)}{3.5}}$
$sin\left(\frac{A}{2}\right)=\frac{1}{\sqrt{5}}$

Answer: Option B. -> 12R2.sin 2A.sin 2B.sin 2C
:
B
Since, area of a triangle
$=\frac{1}{2}\text{(product of the sides)}\times \text{sine of the included angle)}$
$=\frac{1}{2}R\sin 2B\cdot \sin (180°-2A)$
$=\frac{1}{2}{R}^2\sin 2A\cdot \sin 2B\cdot \sin 2C$

Answer: Option B. -> √32  
:
B
In trigonometry one problem can be solved in multiple ways but it is wise to use the best way to get the answer or at least not to use the lengthiest way. Here, we can apply sine rule and get the answer but have a look at the problem again. We are given all the sides. And when sides are given and angles are asked to find ,using cosine rule is a better option.
According to cosine rule -
$cosA=\frac{b^2+c^2-a^2}{2bc}$
$cosA=\frac{(8{)}^2+(7{)}^2-(13{)}^2}{2\left(8\right)\left(7\right)}$ (On putting values of a,b & c)
$cosA=-\frac{1}{2}$
$A=\frac{2\pi }{3}$
$sinA=sin\frac{2\pi }{3}=\frac{\sqrt{3}}{2}$

Answer: Option C. -> 120∘
:
C
We know from trigonometric ratios of half angles that $tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{(s-a)\left(s\right)}}$
Here we need angle opposite to c i.e., C.
Semiperimeter, $S=\frac{a+b+c}{2}=\frac{3+5+7}{2}=\frac{15}{2}$
$tan\frac{C}{2}=\sqrt{\frac{(s-b)(s-a)}{(s-c)\left(s\right)}}$
$=\sqrt{\frac{(s-3)(s-5)}{(s-7)s}}=\sqrt{\frac{(15-6)(15-10)\left(2\right)}{4(15-14)\left(\frac{15}{2}\right)}}$
$=\sqrt{\frac{9\left(5\right)}{15}}=\sqrt{3}\Rightarrow \frac{C}{2}={60}^{\circ }\Rightarrow C={120}^{\circ }$

Answer: Option A. -> √42.18  
:
A
We know the area of a triangle can be calculated using its sides through Heron's formula
$\sqrt{s(s-a)(s-b)(s-c)}$
Where "s" is the semiperimeter of the triangle.
And a, b, c are the sides of it.
So, $s=\frac{3+5+7}{2}=7.5$
And Area = $\sqrt{7.5(7.5-3)(7.5-5)(7.5-7)}$
$=\sqrt{42.18}$

Answer: Option A. -> B = C
:
A
As the median is perpendicular by SAS similarity angle B should be equal to angle C.

Answer: Option A. -> r=Δs,r=2RsinA2sinB2sinC2
:
A
Relation between in radius, Area of triangle and semiperimeter can be given by $r=\frac{\Delta }{s}$ and relation between inradius and circumradius can be given by $r=2Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$

Answer: Option A. -> 10
:
A
We know that the area of a triangle = $\frac{1}{2}$ a.b.sin C
In this C is the angle opposite to the side of length "c" or made in between sides a & b.
Area =$\frac{1}{2}\mathrm{5.8.}sin{30}^{\circ }$
Area = 10 units.

Answer: Option D. -> a=√6, B=105∘, C=15∘
:
D
$A={60}^{\circ }\Rightarrow B+C={120}^{\circ }Weknow,tan\frac{B-C}{2}=\frac{b-c}{b+c}cot\frac{A}{2}\Rightarrow tan\frac{B-C}{2}=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}cot\left(\frac{60}{2}\right)=\frac{2}{2\sqrt{3}}\left(\sqrt{3}\right)=1\therefore \frac{B-C}{2}={45}^{\circ }\Rightarrow B-C={90}^{\circ }B+C={120}^{\circ }\therefore B={105}^{\circ }andC={15}^{\circ },Fromsinerule,a=\frac{bsinA}{sinB}=\sqrt{3}+1\left(\frac{sin60}{sin105}\right)sin105=sin({60}^{\circ }+{45}^{\circ })=\frac{\sqrt{3}}{2}\left(\frac{1}{\sqrt{2}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}a=\frac{(\sqrt{3}+1)\left(\sqrt{3}\right)\left(2\right)\left(\sqrt{2}\right)}{(\sqrt{3}+1)2}=\sqrt{6}\therefore a=\sqrt{6},B={105}^{\circ },C={15}^{\circ }$
Option d is correct.

Answer: Option C. -> Right angled
:
C
Given, $a^2+b^2+c^2=ca+ab\sqrt{3}{a}^2+b^2+c^2-ca-ab\sqrt{3}=0\Rightarrow {(\frac{a\sqrt{3}}{2}-b)}^2+{(\frac{a}{2}-c)}^2=0\Rightarrow a=\frac{2b}{\sqrt{3}}anda=\frac{c}{2}Wecansee,a^2+b^2=c^{2}sinB=\frac{b}{a}=\frac{\sqrt{3}}{2}\ne 1\Rightarrow Notisosceles\therefore GiventriangleisaRightangledtrianglewhichisnotisosceles$.
Option C is correct.