In a triangle ABC,if b=10,a cos^2 frac{C}{2} | vedclass

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

Question

(D) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$. Using $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we get $\frac{a(1+\cos C)}{2} + \frac{c

Vedclass · Accepted Answer

$\frac{5}{\sqrt{3}}$

Vedclass · Answer

(D) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$. Using $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we get $\frac{a(1+\cos C)}{2} + \frac{c(1+\cos A)}{2} = 15$. $\Rightarrow (a+c) + (a \cos C + c \cos A) = 30$. Since $a \cos C + c \cos A = b$,we have $a+c+b = 30$. Given $b=10$,so $a+c = 20$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{30}{2} = 15$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 15\sqrt{3}$. $15\sqrt{3} = \sqrt{15(15-a)(15-10)(15-c)} = \sqrt{75(15-a)(15-c)}$. Squaring both sides: $225 	imes 3 = 75(15-a)(15-c) \Rightarrow 9 = (15-a)(15-c) = 225 - 15(a+c) + ac$. $9 = 225 - 15(20) + ac$ $\Rightarrow 9 = 225 - 300 + ac$ $\Rightarrow ac = 84$. Area $\Delta = \frac{1}{2}ac \sin B = 15\sqrt{3}$ $\Rightarrow \frac{1}{2}(84) \sin B = 15\sqrt{3}$ $\Rightarrow 42 \sin B = 15\sqrt{3}$ $\Rightarrow \sin B = \frac{5\sqrt{3}}{14}$. Then $\cos^2 B = 1 - \sin^2 B = 1 - \frac{75}{196} = \frac{121}{196} \Rightarrow \cos B = \frac{11}{14}$. Finally,$\cot \frac{B}{2} = \sqrt{\frac{1+\cos B}{1-\cos B}} = \sqrt{\frac{1 + 11/14}{1 - 11/14}} = \sqrt{\frac{25/14}{3/14}} = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}$.

In a $\triangle ABC$,if $b=10$,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$,and the area of the triangle is $15\sqrt{3}$ sq. units,then $\cot \frac{B}{2} =$

Similar Questions

Let $S = \{x \in [-\pi, \pi] : \sin x(\sin x + \cos x) = a, a \in Z\}$. Then $n(S)$ is equal to:

If in a triangle $ABC,$ $\cos A \cos B + \sin A \sin B \sin C = 1,$ then the sides are proportional to

The general value of $\theta$ that satisfies both the equations $\cot^3\theta + 3\sqrt{3} = 0$ and $\csc^5\theta + 32 = 0$ is $(n \in I)$.

In a $\triangle ABC$,if $A-B=120^{\circ}$ and $R=8r$,then find the value of $\frac{1+\cos C}{1-\cos C}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module