(B) Using the sine rule,$a = k \sin A, b = k \sin B, c = k \sin C.$Substituting these into the expression:$E = k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos

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(B) Using the sine rule,$a = k \sin A, b = k \sin B, c = k \sin C.$Substituting these into the expression:$E = k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos(C-A) + \sin^3 C \cos(A-B)]$Since $A+B+C = \pi,$ we have $\sin A = \sin(B+C).$Thus,$\sin^3 A \cos(B-C) = \sin^2 A \sin(B+C) \cos(B-C) = \sin^2 A \cdot \frac{1}{2} (\sin 2B + \sin 2C).$Expanding this for all terms and simplifying using the identity $\sin A \sin B \sin C = \frac{abc}{k^3},$ we get:$E = 3k^3 \sin A \sin B \sin C = 3abc.$

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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In a triangle $ABC,$ ${a^3}\cos (B - C) + {b^3}\cos (C - A) + {c^3}\cos (A - B) = $

A
$abc$
B
$3abc$
C
$a + b + c$
D
None of these

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Trigonometrical Equations

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

DifficultTS EAMCET 2008

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The sum of the sides of a right-angled triangle is $42$,and the difference between the median and altitude drawn from the vertex at the right angle is $2$. The area of the triangle is

KVPY 2021

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If in a triangle the angles are in $A.P.$ and $b:c = \sqrt{3}:\sqrt{2}$,then $\angle A$ is equal to .....$^o$.

MediumIIT 1981

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In a triangle $ABC$, if $a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}$, then $b \tan \frac{B}{2} \tan \frac{C-A}{2} =$

MediumTS EAMCET 2022

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In a triangle $ABC$,angle $A$ is greater than angle $B$. If the measures of angles $A$ and $B$ satisfy the equation $3\sin x - 4\sin^3 x - k = 0$ for $0 < k < 1$,then the measure of angle $C$ is:

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In a triangle $ABC,$ ${a^3}\cos (B - C) + {b^3}\cos (C - A) + {c^3}\cos (A - B) = $

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In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

The sum of the sides of a right-angled triangle is $42$,and the difference between the median and altitude drawn from the vertex at the right angle is $2$. The area of the triangle is

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