Given b+c 11 = c+a 12 = a+b 13 for a Δ ABC with usual notation. If A α = B β = C γ , then the ordered triad (α, β, γ) has a value

Correct answer: A. (7,19,25) Let (b+c)/(11)=(c+a)/(12)=(a+b)/(13)=k (Say). b+c=11 k, c+a=12 k, a+b=13 k a+b+c=18 k a=7 k, b=6 k and c=5 k A= 36 k^ 2 +25 k^ 2 -49 k^ 2 2.30 k^ 2 =(1)/(5) and B= 49 k^ 2 +25 k^ 2 -36 k^ 2 2.35 k^ 2 =(19)/(35) and C= 49 k^ 2 +36 k^ 2 -25 k^ 2 2.42 k^ 2 =(5)/(7) A: B: C=7: 19: 25 ( A)/(7)=( B)/(19)=( C)/(25) Hence, required ordered triplet is (7,19,25)

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Properties of Triangles

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Properties of Triangles

Given b+c 11 = c+a 12 = a+b 13 for a Δ ABC with usual notation. If...

Given

\frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}

for a

\Delta \mathrm{ABC}

with usual notation. If

\frac{\cos \mathrm{A}}{\alpha}=\frac{\cos \mathrm{B}}{\beta}=\frac{\cos \mathrm{C}}{\gamma},

then the ordered triad

(\alpha, \beta, \gamma)

has a value

Options

A(7,19,25)
B(3,4,5)
C(5,12,13)
D(19,7,25)

Correct answer

A. (7,19,25)

Step-by-step solution

Let

\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k

(Say).

\therefore \quad b+c=11 k, c+a=12 k, a+b=13 k

\therefore \quad a+b+c=18 k

\therefore \quad a=7 k, b=6 k

and

c=5 k

\therefore \quad \cos A=\frac{36 k^{2}+25 k^{2}-49 k^{2}}{2.30 k^{2}}=\frac{1}{5}

and

\cos B=\frac{49 k^{2}+25 k^{2}-36 k^{2}}{2.35 k^{2}}=\frac{19}{35}

and

\cos C=\frac{49 k^{2}+36 k^{2}-25 k^{2}}{2.42 k^{2}}=\frac{5}{7}

\therefore \quad \cos A: \cos B: \cos C=7: 19: 25

\therefore \quad \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}

Hence, required ordered triplet is (7,19,25)

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