Question

If sines of the angle A and B of a triangle ABC satisfy the equation \[{{c}^{2}}{{x}^{2}}-c\left( a+b \right)x+ab=0\], then the triangle is
$\begin{align}
  & \text{a)}\text{ is acute angled} \\
 & \text{b) is right angled} \\
 & \text{c) is obtuse angled} \\
 & \text{d)}\text{ satisfies }\sin A+\cos A=\dfrac{a+b}{c}\text{ } \\
\end{align}$

Answer 1

Hint: Now we are given that the sum of the roots of the quadratic equation is given by $\dfrac{-b}{a}$. Hence we will use this to form an equation. Now we will simplify the equation using sine rule and hence find the value of angle C. Now we know that the sum of the angles of triangles is $\pi $. Hence using this we will form a relation between angle A and angle B. Now we will substitute the relation in the equation obtained by sum of roots and simplify the equation to find the required solution.

Complete step by step answer:
Now consider the quadratic equation \[{{c}^{2}}{{x}^{2}}-c\left( a+b \right)x+ab=0\]
The equation is a quadratic equation in the form $a{{x}^{2}}+bx+c$
Now we know that the sum of the roots of the equation is given by $\dfrac{-b}{a}$
Now we are given that the roots of the given equation are $\sin A$ and $\sin B$
Hence we get,
$\begin{align}
  & \Rightarrow \sin A+\sin B=\dfrac{c\left( a+b \right)}{{{c}^{2}}} \\
 & \Rightarrow \sin A+\sin B=\dfrac{a+b}{c}.................\left( 1 \right) \\
\end{align}$
Now we know that according to sin rule $\sin A=\dfrac{a}{2R}$. Hence substituting this in equation (1) we get,
$\begin{align}
  & \Rightarrow \dfrac{a}{2R}+\dfrac{b}{2R}=\dfrac{a+b}{c} \\
 & \Rightarrow \dfrac{a+b}{2R}=\dfrac{a+b}{c} \\
 & \Rightarrow c=2R \\
\end{align}$
Now again using sin rule we get,
$\begin{align}
  & \Rightarrow 2R\sin C=2R \\
 & \Rightarrow \sin C=1 \\
\end{align}$
Hence we get that the value of C is $\dfrac{\pi }{2}$
Hence now we know that the triangle is a right angle triangle.
Now we know that the sum of angles of triangle is equal to $\pi $
Hence we get,
$\begin{align}
  & \Rightarrow \angle A+\angle B+\angle C=\pi \\
 & \Rightarrow \angle A+\angle B+\dfrac{\pi }{2}=\pi \\
 & \Rightarrow \angle A+\angle B=\dfrac{\pi }{2} \\
\end{align}$
Hence we can say that $\angle B=\dfrac{\pi }{2}-\angle A$
Now substituting the value in equation we get,
$\Rightarrow \sin A+\sin \left( \dfrac{\pi }{2}-\angle A \right)=\dfrac{a+b}{c}$
Now we know that $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $. Hence using this identity we get,
$\Rightarrow \sin A+\cos A=\dfrac{a+b}{c}$

So, the correct answer is “Option b and d”.

Note: Now note that for a quadratic equation we also have that the product of the roots is given by $\dfrac{c}{a}$. Also note that when using sin rule the side a is the side opposite to angle A. Similarly we have b and c are the sides opposite to angle B and angle C.