SATHEE: Properties of Triangles 1 Question 2

Properties of Triangles 1 Question 2

2. Given, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $△ A B C$ with usual notation. If $\frac{\cos A}{α} = \frac{\cos B}{β} = \frac{\cos C}{γ}$ , then the ordered $triad (α, β, γ)$ has a value

(2019 Main, 11 Jan II)

(a) $(19, 7, 25)$

(b) $(3, 4, 5)$

(d) $(7, 19, 25)$

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Answer:

Correct Answer: 2. (d)

Solution:

Given, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = λ$ (say)

$\begin{aligned} b + c = 11 λ, c + a = 12 λ and a + b = 13 λ \\ \Rightarrow 2 (a + b + c) = 36 λ \\ \Rightarrow a + b + c = 18 λ \end{aligned}$

From Eqs. (i) and (ii), we get

$a = 7 λ, b = 6 λ, c = 5 λ$

Now,

$\begin{aligned} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{λ^{2} [36 + 25 - 49]}{60 λ^{2}} = \frac{12}{60} = \frac{1}{5} \\ \begin{aligned} \cos B = & \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{λ^{2} [49 + 25 - 36]}{70 λ^{2}} = \frac{19}{35} \\ and \cos C & = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{λ^{2} [49 + 36 - 25]}{84 λ^{2}} \\ = \frac{60}{84} = \frac{5}{7} \end{aligned} \\ Thus, \cos A = \frac{1}{5} = \frac{7}{35}, \\ \cos B = \frac{19}{35}, \cos C = \frac{25}{35} \\ \frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25} = \frac{1}{35} \\ \Rightarrow (α, β, γ) = (7, 19, 25) \end{aligned}$

Properties of Triangles 1 Question 2

2. Given, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $△ A B C$ with usual notation. If $\frac{\cos A}{α} = \frac{\cos B}{β} = \frac{\cos C}{γ}$ , then the ordered $triad (α, β, γ)$ has a value

Answer:

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Properties of Triangles 1 Question 2

2. Given, b+c11=c+a12=a+b13 for a △ABC with usual notation. If cos⁡Aα=cos⁡Bβ=cos⁡Cγ, then the ordered triad(α,β,γ) has a value

Answer:

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2. Given, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $△ A B C$ with usual notation. If $\frac{\cos A}{α} = \frac{\cos B}{β} = \frac{\cos C}{γ}$ , then the ordered $triad (α, β, γ)$ has a value