⇒Trigonometric Functions

Correct answer option is (D)

Here, $\mathrm{A}(-2,3),\mathrm{B}(6,-1),\mathrm{C}(4,3)$ are the vertices of $\mathrm{△}ABC$.

Let $\mathrm{F}$ be the circumcentre of $\mathrm{△}\mathrm{ABC}$.

Let $FD$ and $FE$ be the perpendicular bisectors of the sides $\mathrm{BC}$ and $\mathrm{AC}$ respectively.

$\therefore \mathrm{D}$ and $\mathrm{E}$ are the midpoints of side $\mathrm{BC}$ and $\mathrm{AC}$ respectively.

$\begin{array}{ll}\therefore & D\equiv (\frac{6+4}{2},\frac{-1+3}{2})\\ \therefore & D=(5,1)\\ & \text{ and }E\equiv (\frac{-2+4}{2},\frac{3+3}{2})\\ & \therefore \mathrm{E}=(1,3)\end{array}$

Now, slope of $BC=\frac{3-(-1)}{4-6}=\frac{4}{-2}=-2$

$\therefore$ Slope of $\mathrm{FD}=\frac{1}{2}\dots [\because \mathrm{FD}\perp \mathrm{BC}]$

Since FD passes through $(5,1)$ and has slope $\frac{1}{2}$, equation of $\mathrm{FD}$ is

$\begin{array}{rl}& y-1=\frac{1}{2}(x-5)\\ \therefore & 2(y-1)=x-5\\ \therefore & 2y-2=x-5\\ \therefore & x-2y-3=0\text{.... (i)}\end{array}$

Since both the points $\mathrm{A}$ and $\mathrm{C}$ have same $y$ co-ordinates i.e. 3, the given points lie on the line $y=3$.

Since the equation $\mathrm{FE}$ passes through $\mathrm{E}(1,3)$, the equation of $\mathrm{FE}$ is $x=1$. ..... (ii)

To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).

Substituting the value of $x$ in (i), we get

$\begin{array}{rl}& 1-2y-3=0\\ \therefore & y=-1\end{array}$

$\therefore$ Co-ordinates of circumcentre $\mathrm{F}\equiv (1,-1)$.

Correct answer option is (D)

By cosine rule, we get

$\begin{array}{rl}{a}^2& =b^2+c^2-2bc\cos A\\ & =(\sqrt{3}{)}^2+(1{)}^2-2(\sqrt{3})(1)\cos \left({30}^{\circ }\right)\\ & =3+1-2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\\ & =4-3\\ & =1\end{array}$

$\therefore \mathrm{a}=1$

$\therefore$ Largest angle is angle $\mathrm{B}$

$\begin{array}{rl}& \cos B=\frac{{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2}{2\mathrm{ca}}=\frac{1+1-3}{2\times 1\times 1}=\frac{-1}{2}\\ & \therefore B={120}^{\circ }\end{array}$

Correct answer option is (A)

Let the angles of the triangle be $4x,x$ and $x$.

$\begin{array}{rl}\therefore & 4x+x+x={180}^{\circ }\Rightarrow 6x={180}^{\circ }\Rightarrow x={30}^{\circ }\\ & \frac{\sin {120}^{\circ }}{a}=\frac{\sin {30}^{\circ }}{b}=\frac{\sin {30}^{\circ }}{c}\\ \therefore & a:(a+b+c)\\ & =(\sin {120}^{\circ }):(\sin {120}^{\circ }+\sin {30}^{\circ }+\sin {30}^{\circ })\\ & =\frac{\sqrt{3}}{2}:\frac{\sqrt{3}+2}{2}=\sqrt{3}:(\sqrt{3}+2)\end{array}$

Correct answer option is (C)

$\begin{array}{rl}& a\cdot {\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=\frac{3b}{2}\\ & \Rightarrow a\left(\frac{1+\cos C}{2}\right)+c\left(\frac{1+\cos A}{2}\right)=\frac{3b}{2}\\ & \Rightarrow a+a\cos C+c+c\cos A=3b\\ & \Rightarrow a+b+c=3b\dots [\because b=c\cos A+a\cos C]\\ & \Rightarrow a+c=2b\end{array}$

$\therefore a,b,c$ are in A.P.

Correct answer option is (B)

By sine rule, we get

$\begin{array}{rl}& \frac{\sin A}{a}=\frac{\sin B}{b}\\ & \Rightarrow \frac{\sin {60}^{\circ }}{6}=\frac{x}{8}\dots [\because B={\sin }^{-1}(x)]\\ & \Rightarrow x=\frac{\sqrt{3}}{2}\times \frac{8}{6}\\ & \Rightarrow x=\frac{2}{\sqrt{3}}\end{array}$