(B) Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Sub

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(B) Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting these into the expression:$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}$$= \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{2abc}$$= \frac{a^2 + b^2 + c^2}{2abc}$.

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

Easy

In $\Delta ABC$,$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

A
$\frac{a^2 + b^2 + c^2}{abc}$
B
$\frac{a^2 + b^2 + c^2}{2abc}$
C
$\frac{2(a^2 + b^2 + c^2)}{abc}$
D
$a^2 + b^2 + c^2$

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Class 11

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

In a triangle $ABC$,if $a \neq b$,then the value of $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$ is:

DifficultAP EAMCET 2022

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In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

DifficultTS EAMCET 2013

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In a right-angled $\triangle ABC$, the measures of the angles are in an Arithmetic Progression ($A$.$P$.). If its smallest side is $4 \text{ units}$, then the area of $\triangle ABC$ is:

DifficultMHT CET 2026

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In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

EasyAP EAMCET 2017

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In $\triangle ABC$,$\angle B = \frac{\pi}{4}$ and $\angle C = \frac{\pi}{3}$. If the area of the triangle is $54 + 18\sqrt{3}$ sq. units,then $a =$

MediumTS EAMCET 2021

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In $\Delta ABC$,$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

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In a triangle $ABC$,if $a \neq b$,then the value of $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$ is:

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

In a right-angled $\triangle ABC$, the measures of the angles are in an Arithmetic Progression ($A$.$P$.). If its smallest side is $4 \text{ units}$, then the area of $\triangle ABC$ is:

In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

In $\triangle ABC$,$\angle B = \frac{\pi}{4}$ and $\angle C = \frac{\pi}{3}$. If the area of the triangle is $54 + 18\sqrt{3}$ sq. units,then $a =$

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