In Delta ABC,frac{sin(A - B)}{sin(A + B)} = | vedclass

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

Question

(A) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\sin(A + B) = \sin(180^{\circ} - C) = \sin C$. Thus,$\frac{\sin(A - B)}{\sin(A + B)} =

Vedclass · Accepted Answer

$\frac{a^2 - b^2}{c^2}$

Vedclass · Answer

(A) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\sin(A + B) = \sin(180^{\circ} - C) = \sin C$. Thus,$\frac{\sin(A - B)}{\sin(A + B)} = \frac{\sin A \cos B - \sin B \cos A}{\sin C}$. Using the sine rule,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$. Substituting these,we get $\frac{\frac{a}{2R} \cos B - \frac{b}{2R} \cos A}{\frac{c}{2R}} = \frac{a \cos B - b \cos A}{c}$. Using the projection formula,$a = b \cos C + c \cos B$ and $b = a \cos C + c \cos A$,we have $a \cos B - b \cos A = (b \cos C + c \cos B) \cos B - (a \cos C + c \cos A) \cos A$. Alternatively,using the cosine rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting these: $\frac{a(\frac{a^2 + c^2 - b^2}{2ac}) - b(\frac{b^2 + c^2 - a^2}{2bc})}{c} = \frac{\frac{a^2 + c^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c}}{c} = \frac{a^2 + c^2 - b^2 - b^2 - c^2 + a^2}{2c^2} = \frac{2a^2 - 2b^2}{2c^2} = \frac{a^2 - b^2}{c^2}$.

In $\Delta ABC$,$\frac{\sin(A - B)}{\sin(A + B)} = $

Similar Questions

In a $\triangle ABC$,if $a=5, b=6, c=7$,then the length of the median drawn from $B$ is

If $a = 2, b = 3, c = 5$ in $\Delta ABC$,then $C = $

In a triangle $ABC$ with usual notations,if $a=13, b=14, c=15$,then $\sin A=$

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

Sides of a triangle are $2 \ cm$,$\sqrt{6} \ cm$,and $(\sqrt{3} + 1) \ cm$. The smallest angle of the triangle is .....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module