(C) Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Given $\angle C = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c

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(C) Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Given $\angle C = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.This implies $ab = a^2+b^2-c^2$,or $a^2+b^2 = ab+c^2$.Now,substitute $a^2+b^2$ in the given expression:$\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)} = \frac{ca+cb+ab+c^2}{bc+ab+c^2+ac}$.Since the numerator and denominator are identical,the value is $1$.

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

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Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $\angle C=60^{\circ}$,then the value of $\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)}$ is

AP EAMCET 2020 All AP EAMCET

A
$\frac{1}{2}$
B
$\frac{\sqrt{3}}{2}$
C
$1$
D
$\sqrt{3}$

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Relation between sides and angles, Solutions of triangles

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Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $\angle C=60^{\circ}$,then the value of $\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)}$ is

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