SATHEE: Properties of Triangles 1 Question 22

Properties of Triangles 1 Question 22

22. With usual notation, if in a $△ A B C \frac{b + c}{11} = \frac{c + a}{12}$ $= \frac{a + b}{13}$ , then prove that $\frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25}$ .

(1984, 4M)

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Solution:

Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = λ$

$\Rightarrow (b + c) = 11 λ, c + a = 12 λ, a + b = 13 λ$

$\Rightarrow 2 (a + b + c) = 36 λ$

$\Rightarrow a + b + c = 18 λ$

On solving Eqs. (i) and (ii), we get

$\begin{matrix} a = 7 λ, b = 6 λ and c = 5 λ \\ \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{36 λ^{2} + 25 λ^{2} - 49 λ^{2}}{2 (30) λ^{2}} = \frac{1}{5} \\ \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{49 λ^{2} + 25 λ^{2} - 36 λ^{2}}{70 λ^{2}} = \frac{19}{35} \\ \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{49 λ^{2} + 36 λ^{2} - 25 λ^{2}}{84 λ^{2}} = \frac{5}{7} \end{matrix}$

$∴ \cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{5}{7} = 7 : 19 : 25$

Properties of Triangles 1 Question 22

22. With usual notation, if in a $△ A B C \frac{b + c}{11} = \frac{c + a}{12}$ $= \frac{a + b}{13}$ , then prove that $\frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25}$ .

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Properties of Triangles 1 Question 22

22. With usual notation, if in a △ABCb+c11=c+a12 =a+b13, then prove that cos⁡A7=cos⁡B19=cos⁡C25.

Solution:

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22. With usual notation, if in a $△ A B C \frac{b + c}{11} = \frac{c + a}{12}$ $= \frac{a + b}{13}$ , then prove that $\frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25}$ .