Chapter 8.2, Problem 60AYU

To determine

To find: To Derive the formula $a=b\cos C+c\cos B$ for any triangle.

Answer to Problem 60AYU

The formula is

  $a=b\cos C+c\cos B$ ,

Explanation of Solution

Given information:

The given triangle is ABC

Its sides and opposite angles are $a,b,c$ , $A,B,C$

Concept used:

Law of Sine:

If ABC is a triangle with sides $a$ , $b$ , $c$ then

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

The given triangle is ABC

Its sides and opposite angles are $a,b,c$ , $A,B,C$

By using the law of Sine to take the constant is $k$ ,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

The equations are,

  $\begin{array}{l}\sin A=ak\mathrm{.....}\left(1\right)\\ \sin B=bk\mathrm{.....}\left(2\right)\\ \sin C=ck\mathrm{.....}\left(3\right)\end{array}$

The sum of angle of the triangle $ABC$ is,

  $\begin{array}{l}A+B+C={180}^{\circ }\\ A=\left[{180}^{\circ }-\left(B+C\right)\right]\end{array}$

Multiplying with sine of both sides

  $\sin A=\sin \left[{180}^{\circ }-\left(B+C\right)\right]$

The angle of $\sin \left({180}^{\circ }-\theta \right)=\sin \theta$

  $\sin A=\sin \left(B+C\right)$

By using sum to product formula,

  $\begin{array}{l}\sin A=\sin \left(B+C\right)\\ \sin A=\sin B.\cos C+\cos B.\sin C\end{array}$

From equations $\mathrm{....}\left(1\right),\mathrm{..}\left(2\right)$ and $\mathrm{..}\left(3\right)$

  $\begin{array}{l}\sin A=\sin B.\cos C+\cos B.\sin C\\ ak=bk.\cos C+\cos B.ck\\ a=b.\cos C+c.\cos B\end{array}$

Hence, the formula is $a=b.\cos C+c.\cos B$ .