In Delta ABC, if a = 16, b = 24 and c = 20, t | vedclass

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

Question

(A) Given sides are $a = 16, b = 24, c = 20$.Semi-perimeter $s = \frac{a + b + c}{2} = \frac{16 + 24 + 20}{2} = \frac{60}{2} = 30$.Using the formula $

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(A) Given sides are $a = 16, b = 24, c = 20$.Semi-perimeter $s = \frac{a + b + c}{2} = \frac{16 + 24 + 20}{2} = \frac{60}{2} = 30$.Using the formula $\cos \frac{B}{2} = \sqrt{\frac{s(s - b)}{ac}}$:$\cos \frac{B}{2} = \sqrt{\frac{30(30 - 24)}{16 	imes 20}}$$\cos \frac{B}{2} = \sqrt{\frac{30 	imes 6}{320}}$$\cos \frac{B}{2} = \sqrt{\frac{180}{320}} = \sqrt{\frac{18}{32}} = \sqrt{\frac{9}{16}}$$\cos \frac{B}{2} = \frac{3}{4}$.

In $\Delta ABC,$ if $a = 16, b = 24$ and $c = 20,$ then $\cos \frac{B}{2} = $

Similar Questions

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is

In a triangle $ABC$ with usual notations,find the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$.

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

If in a triangle $ABC$,$\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then the value of the angle $A$ is:

If $b+c=3a$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module