In $△ A B C, b c = 2 b^{2} c o s A + 2 c^{2} c o s A - 4 b c c o s^{2} A$ , then $△ A B C$ is

isosceles but not necessarily equilaterial

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

equilaterial

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

right angled but not necessarily isosceles

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

right angles isosceles

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

Open in App

Solution

The correct option is A isosceles but not necessarily equilaterial
$[b c = 2 b^{2} cos A + 2 c^{2} cos A + 4 b c {cos}^{2} A]$
$[(2 b cos A - c) (b - 2 c cos A)]$
$[cos A = \frac{c}{2 b}, \frac{b}{2 c}]$
$[cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}]$
Solving for the solutions of $cos A$ we get that the triangle is isosceles and not necessarily equilateral.

Suggest Corrections

Similar questions

In ∆ABC and ∆PQR, if AB = AC, ∠C = ∠P and ∠B = ∠Q, then the two triangles are:

Q. If

\frac{cos A + 2 cos C}{cos A + 2 cos B} = \frac{sin B}{sin C}

the the triangle

A B C

is either isosceles or right angled.

$I n Δ A B C a n d Δ P Q R, it is given that AB = AC, ∠ C = ∠ P a n d ∠ B = ∠ Q . Then, the two triangles are (a) isosceles but not congruent (b) isosceles and congruent (c) congruent but not isosceles (d) neither congruent nor isosceles$

Q. If in a

△ A B C,

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

then

△ A B C

Q. Question 10
In

Δ A B C

and

Δ P Q R

, if

A B = A C, ∠ C = ∠ P

and

∠ B = ∠ Q

. The two triangles are
(A) isosceles but not congruent
(B) isosceles and congruent
(C) congruent but not isosceles
(D) neither congruent nor isosceles

Join BYJU'S Learning Program

In △ABC,bc=2b2cosA+2c2cosA−4bccos2A, then △ABC is

The correct option is A isosceles but not necessarily equilaterial[bc=2b2cosA+2c2cosA+4bccos2A][(2bcosA−c)(b−2ccosA)][cosA=c2b,b2c][cosA=b2+c2−a22bc]Solving for the solutions of cosA we get that the triangle is isosceles and not necessarily equilateral.

In $△ A B C, b c = 2 b^{2} c o s A + 2 c^{2} c o s A - 4 b c c o s^{2} A$ , then $△ A B C$ is