In a triangle ABC,sin A and sin B satisfy the | vedclass

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

Question

(D) Given the quadratic equation $c^2 x^2 - c(a+b)x + ab = 0$. Factoring the equation: $c^2 x^2 - cax - cbx + ab = 0$ $cx(cx - a) - b(cx - a) = 0$ $(c

Vedclass · Accepted Answer

$\sin A + \cos A = \frac{a+b}{c}$

Vedclass · Answer

(D) Given the quadratic equation $c^2 x^2 - c(a+b)x + ab = 0$. Factoring the equation: $c^2 x^2 - cax - cbx + ab = 0$ $cx(cx - a) - b(cx - a) = 0$ $(cx - a)(cx - b) = 0$ Thus,the roots are $x = \frac{a}{c}$ and $x = \frac{b}{c}$. Since $\sin A$ and $\sin B$ are the roots,we have $\sin A = \frac{a}{c}$ and $\sin B = \frac{b}{c}$. From the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Substituting $\sin A = \frac{a}{c}$ and $\sin B = \frac{b}{c}$,we get $c = \frac{a}{\sin A}$ and $c = \frac{b}{\sin B}$. This implies $\frac{a}{\sin A} = c$ and $\frac{b}{\sin B} = c$. Comparing with the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$,we get $\frac{c}{\sin C} = c$,which means $\sin C = 1$. Therefore,$\angle C = 90^\circ$,so the triangle is right-angled at $C$. In a right-angled triangle at $C$,$\sin A = \frac{a}{c}$ and $\cos A = \sin B = \frac{b}{c}$. Thus,$\sin A + \cos A = \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$.

In a $\triangle ABC$,$\sin A$ and $\sin B$ satisfy the equation $c^2 x^2 - c(a+b)x + ab = 0$. Then:

Similar Questions

If the reciprocals of the lengths of the sides of a $\triangle ABC$ are in harmonic progression,then its ex-radii $r_1, r_2, r_3$ are in

The upper $\frac{3}{4}$th portion of a vertical pole subtends an angle $\tan^{-1}\left(\frac{3}{5}\right)$ at a point in the horizontal plane through its foot and at a distance $40 \ m$ from the foot. $A$ possible height of the vertical pole is $....... \ m$.

In a $\triangle ABC$,if $\cos A + \cos B + \cos C = a + b \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,then $(a + b)$ is equal to

The general solution of the equation $\sqrt{6-5 \cos x+7 \sin ^2 x}-\cos x=0$ also satisfies the equation

In any triangle $ABC$,find the value of $\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module