In any triangle ABC,a(b cos C - c cos B) equa | vedclass

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

Question

(B) Using the projection formula or the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting t

Vedclass · Accepted Answer

$b^2 - c^2$

Vedclass · Answer

(B) Using the projection formula or the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting these into the expression: $a(b \cos C - c \cos B) = ab \cos C - ac \cos B$ $= ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right)$ $= \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2}$ $= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$ $= \frac{2b^2 - 2c^2}{2}$ $= b^2 - c^2$.

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

Similar Questions

In a triangle $ABC$,with usual notations,$2 ac \sin \left(\frac{A-B+C}{2}\right)$ is equal to

With usual notations,if the angles of a triangle are in the ratio $1: 2: 3$,then their corresponding sides are in the ratio.

Evaluate $\cot \left( \frac{A + B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right)$ in terms of sides $a$ and $b$.

If the area of triangle $ABC$ is $b^2-(c-a)^2$,then $\tan B=$

The angles of a triangle are in $A.P.$ and the greatest angle is double the least angle. Find the sine of the third angle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module