(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Rearranging the terms,we get the quadratic equation in $b$: $b^2 - (2c \cos A)b +

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(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Rearranging the terms,we get the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c^2 - a^2) = 0$.Since $c \sin A According to the properties of quadratic equations,the sum of the roots is given by the negative of the coefficient of $b$ divided by the coefficient of $b^2$.Therefore,$b_1 + b_2 = -(-(2c \cos A)) / 1 = 2c \cos A$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Medium

If in a triangle $ABC$,$a, b, c$ are sides and angle $A$ is given,and $c \sin A < a < c$,and $b_1$ and $b_2$ are two possible values of $b$,then:

A
$b_1 + b_2 = 2c \cos A$
B
$b_1 + b_2 = c \cos A$
C
$b_1 + b_2 = 3c \cos A$
D
$b_1 + b_2 = 4c \sin A$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

With usual notations,if triangle $ABC$ is right-angled at $C$,then $\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right) \sin (A-B) =$

MediumMHT CET 2020

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The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

Easy

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In a triangle $ABC$, with usual notations, if $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ and $\lambda \tan^2 \frac{A}{2} = 455$, then $\lambda = $

DifficultMHT CET 2026

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If in triangle $ABC$,$\cos A = \frac{\sin B}{2\sin C}$,then the triangle is

Easy

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Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.

KVPY 2020

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If in a triangle $ABC$,$a, b, c$ are sides and angle $A$ is given,and $c \sin A < a < c$,and $b_1$ and $b_2$ are two possible values of $b$,then:

Similar Questions

With usual notations,if triangle $ABC$ is right-angled at $C$,then $\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right) \sin (A-B) =$

The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

In a triangle $ABC$, with usual notations, if $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ and $\lambda \tan^2 \frac{A}{2} = 455$, then $\lambda = $

If in triangle $ABC$,$\cos A = \frac{\sin B}{2\sin C}$,then the triangle is

Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.

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