Let ΔABC is 11b+c=12c+a=12a+b (where AB=c,BC=aandAC=b) and αcosA=βcosB=γcosC then possible ordered tripblet of (α,β,γ) is (A) (9,17,25) (B) (19,7,25) (C) (7,19,25) (D) (19,25,7)

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Let {A}{B}{C} is {{{b}+{c}}}{{11}}={{{c}+{a}}}{{12}}={{{a}+{b}

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Let $Δ A B C$ is $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{12}$ (where $A B = c, B C = a and A C = b$ ) and $\frac{cos A}{α} = \frac{cos B}{β} = \frac{cos C}{γ}$ then possible ordered tripblet of $(α, β, γ)$ is (A) $(9, 17, 25)$ (B) $(19, 7, 25)$ (C) $(7, 19, 25)$ (D) $(19, 25, 7)$

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Let

Δ A B C

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

(where

A B = c, B C = a and A C = b

) and

\frac{cos A}{α} = \frac{cos B}{β} = \frac{cos C}{γ}

then possible ordered tripblet of

(α, β, γ)

is (A)

(9, 17, 25)

(B)

(19, 7, 25)

(C)

(7, 19, 25)

(D)

(19, 25, 7)

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Question Text	Let $Δ A B C$ is $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{12}$ (where $A B = c, B C = a and A C = b$ ) and $\frac{cos A}{α} = \frac{cos B}{β} = \frac{cos C}{γ}$ then possible ordered tripblet of $(α, β, γ)$ is (A) $(9, 17, 25)$ (B) $(19, 7, 25)$ (C) $(7, 19, 25)$ (D) $(19, 25, 7)$
Answer Type	Video solution: 1
Upvotes	150