If in Delta ABC, a = 6, b = 3 and cos(A - B) | vedclass

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

Question

(C) Given: $a = 6, b = 3, \cos(A - B) = \frac{4}{5}.$Using the formula $\cos(A - B) = \frac{1 - 	an^2(\frac{A - B}{2})}{1 + 	an^2(\frac{A - B}{2})},

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given: $a = 6, b = 3, \cos(A - B) = \frac{4}{5}.$Using the formula $\cos(A - B) = \frac{1 - 	an^2(\frac{A - B}{2})}{1 + 	an^2(\frac{A - B}{2})},$$\frac{4}{5} = \frac{1 - t^2}{1 + t^2} \implies 4 + 4t^2 = 5 - 5t^2 \implies 9t^2 = 1 \implies t = \frac{1}{3}.$So,$	an(\frac{A - B}{2}) = \frac{1}{3}.$Using Napier's Analogy: $	an(\frac{A - B}{2}) = \frac{a - b}{a + b} \cot(\frac{C}{2}).$$\frac{1}{3} = \frac{6 - 3}{6 + 3} \cot(\frac{C}{2}) = \frac{3}{9} \cot(\frac{C}{2}) = \frac{1}{3} \cot(\frac{C}{2}).$$\cot(\frac{C}{2}) = 1 \implies \frac{C}{2} = 45^\circ \implies C = 90^\circ.$The area of $\Delta ABC = \frac{1}{2} ab \sin C = \frac{1}{2} 	imes 6 	imes 3 	imes \sin 90^\circ = 9 	imes 1 = 9 \, square \, unit.$

If in $\Delta ABC,$ $a = 6, b = 3$ and $\cos(A - B) = \frac{4}{5},$ then its area will be ..... $square \, unit.$

Similar Questions

If $x = \frac{n\pi}{2}$ satisfies the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$ and the inequality $\left| \frac{x}{2} - \frac{\pi}{2} \right| \le \frac{3\pi}{4}$,then:

In a $\triangle ABC$,if $\cos A + \cos B + \cos C = a + b \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,then $(a + b)$ is equal to

If $\alpha, \beta, \gamma$ are angles of a triangle,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$ is

Let $ABC$ be a triangle and let $D$ be the mid-point of $BC$. Suppose $\cot (\angle CAD) : \cot (\angle BAD) = 2 : 1$. If $G$ is the centroid of $\triangle ABC$,then the measure of $\angle BGA$ is (in $^{\circ}$)

In a triangle $PQR$ with usual notations,$\angle R = \frac{\pi}{2}$. If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module