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TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 26 of 160

Marks: +1, -0

\frac{a^3+b^3+c^3}{\sin^3 A + \sin^3 B + \sin^3 C}=8

, then the maximum value of c is

Solution:

Given that,

\; \frac{a^{3}+b^{3}+c^{3}}{\sin^{3} A+\sin^{3} B+\sin^{3} C}=8 \;\; \ldots

(i)

Using sine rule for a triangle

A\;B\;C

\; \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

{where

2R \rightarrow

circumradius }

\therefore a = 2R \sin A, b = 2R \sin B, c = 2R \sin C

Substituting the values of

a, b, c

in Eq .(i), we get

\; \frac{(2R \sin A)^3 + (2R \sin B)^3 + (2R \sin C)^3}{\sin^3 A + \sin^3 B + \sin^3 C} = 8

\; \frac{(2R)^3 (\sin^3 A + \sin^3 B + \sin^3 C)}{\sin^3 A + \sin^3 B + \sin^3 C} = 8

(2R)^3 = 8

2R = 2

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