Question

If the sides of a triangle are $3$, $5$, and $7$ and the smallest and greatest angles are $\alpha$ and $\beta$ respectively, then $4(1-\cos \alpha )(1-\cos \beta )$ is equal to:

• A. $\frac{3}{5}$
• B. $\frac{3}{7}$
• C. $\frac{3}{4}$
• D. $\frac{3}{11}$

Answer 1

The correct option is B

$\frac{3}{7}$

Explanation of correct answer.

It is given that,

$$\begin{gathered} a=3 \\ b=5 \\ c=7 \end{gathered}$$

Therefore,

$\begin{array}{rcl}\cos A& =& \cos \alpha \\ & =& \frac{\left(b^2+c^2-a^2\right)}{2bc}\\ & =& \frac{\left(5^2+7^2-3^2\right)}{2\times 5\times 7}\\ & =& \frac{13}{14}\end{array}$

Again,

$$\begin{gathered} \cos C=\cos \beta \\ =\frac{\left(a^2+b^2-c^2\right)}{2ab} \\ =\frac{\left(3^2+5^2-7^2\right)}{2\times 3\times 5} \\ =-\frac{1}{2} \end{gathered}$$

Now,

$$\begin{gathered} 4(1-\cos \alpha )(1-\cos \beta )=4(1-\frac{13}{14})(1+\frac{1}{2}) \\ =\frac{3}{7} \end{gathered}$$

Hence, Option (B) is correct.