Question

If two sides a, b and angle A be such that two triangles are formed, then the sum of two values of the third side is
A. $2b\sin A$
B. $2b\cos A$
C. ${\displaystyle \frac{b}{a}}\cos A$
D. $\left(c+b\right)\cos A$

Answer 1

Hint:In the above question, we are provided with a triangle with sides a, b and c with angle A, B and C. These angles are taken according to the sides like-ways we have to find the sum of values of the third side so, use angle A and use the cosine formula. then form the quadratic equation, using the properties of the quadratic equation finds the answer.

Complete step by step solution:
In the above question, we had a triangle with sides namely $a,b,c$ having the angle A. and we had to find the sum of two values of the third side. for that using the cosine formula,
$\cos A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}$
Now, cross multiplying and forming a quadratic equation,
$2bc.\cos A=b^2+c^2-a^2$
Now, forming a quadratic equation in terms of c
$c^2-(2b.\cos A)c+(b^2-a^2)=0$
Let $c_1,c_2$ be the roots of the above equations
And we know that the sum of roots of a quadratic equation is equal to the coefficient of the “c” and the product of the roots of the quadratic equation is equal to the product of the coefficient of the constant term and the coefficient of the $c^2$
Hence, on the behalf of the above statement
$c_1+c_2=2b.\cos A$
And $c_1{c}_2=b^2-a^2$
Hence, the sum of the values of the third side is $2b\cos A$
So, the correct option is B.

Note: Remember to use the cosine formula carefully, while writing the cosine formula for angle A which is between the sides b and c. So, the squares of sides b and c are taken as positive and a as negative with dividing the product of sides b and c.