If sines of the angles A and B of a triangle ABC satisfy the equation $c^{2} x^{2} - c (a + b) x + a b = 0$ , then the triangle

is acute angled

No worries! We‘ve got your back. Try BYJU‘S free classes today!

is right angled

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

is obtuse angled

No worries! We‘ve got your back. Try BYJU‘S free classes today!

satisfies sin A + cos A

=

(a + b)/c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B is right angled
D satisfies sin A + cos A $=$ (a + b)/c
Sum of roots $s i n A + s i n B$ $= \frac{a + b}{c}$
$= \frac{s i n A + s i n B}{s i n C}$
$s i n (A) + s i n (B) [1 - \frac{1}{s i n C}] = 0$
$s i n C = 1$
$C = 90^{0}$ .
Hence,
$s i n A + c o s A$
$= s i n A + s i n B$
$= \frac{a s i n A}{a} + \frac{b s i n B}{b}$
$= \frac{s i n C (a + b)}{c}$
$= \frac{a + b}{c}$
$= \frac{a + b}{c}$

Suggest Corrections

Similar questions

If $s i n A$ and $s i n B$ of a $Δ A B C$ satisfy $c^{2} x^{2} - c (a + b) x + a b = 0$ then the triangle is

sin A

sin B

A B C

c^{2} x^{2} - c (a + b) x + a b = 0

Δ A B C, sin C + cos C + sin (2 B + C) - cos (2 B + C) = 2 \sqrt{2} .

Δ A B C

△ A B C

c^{3} x^{3} - c^{2} (a + b + c) x^{2} + l x + m = 0

a, b, c

△ A B C

△ A B C

If for a $△ A B C, c o t A . c o t B . c o t C > 0$ then the triangle is

Join BYJU'S Learning Program