If in the Delta ABC,AB = 2BC,then find the ra | vedclass

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

Question

(D) Given $AB = c$ and $BC = a$. Since $AB = 2BC$,we have $c = 2a$.Using the Napier's Analogy,we know that $	an \frac{C - A}{2} = \frac{c - a}{c + a}

Vedclass · Accepted Answer

$1:3$

Vedclass · Answer

(D) Given $AB = c$ and $BC = a$. Since $AB = 2BC$,we have $c = 2a$.Using the Napier's Analogy,we know that $	an \frac{C - A}{2} = \frac{c - a}{c + a} \cot \frac{B}{2}$.Rearranging this,we get $\frac{	an (B/2)}{\cot ((C - A)/2)} = 	an \frac{B}{2} 	an \frac{C - A}{2}$.Substituting the expression for $	an \frac{C - A}{2}$,we get $	an \frac{B}{2} \left( \frac{c - a}{c + a} \cot \frac{B}{2} ight) = \frac{c - a}{c + a}$.Substituting $c = 2a$,we get $\frac{2a - a}{2a + a} = \frac{a}{3a} = \frac{1}{3}$.Thus,the ratio is $1:3$.

If in the $\Delta ABC$,$AB = 2BC$,then find the ratio $\tan \frac{B}{2} : \cot \left( \frac{C - A}{2} \right)$.

Similar Questions

In $\Delta ABC,$ if ${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3},$ then the triangle is:

Let $ABC$ be an acute-angled triangle with area $R$. Then,$\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2} = $

In $\triangle ABC$,$a^2 \sin 2B + b^2 \sin 2A =$

In $\triangle ABC$,$\tan \frac{A}{2} + \tan \frac{B}{2} =$

In $\triangle ABC$,what is the value of $\cos A + \cos B + \cos C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module