In - ABC- b-c11- c-a12- a-b13- then sin A-sin B-sin C-

Name: JEE- Grade 12- Mathematics- Solutions of Triangle- Session 01- W14
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let nbsp; b+c11= c+a12= a+b13 = k; ⇒ b+c = 11k ⋯ (i) ⇒ c+a = 12k ⋯ (ii) ⇒ a+b = 13k ⋯ (iii) Adding (i),(ii),(iii) and dividing by 2 ⇒ a+b+c = 18k ∴ a=7k ...

Let nbsp; b+c11= c+a12= a+b13 = k; ⇒ b+c = 11k ⋯ (i) ⇒ c+a = 12k ⋯ (ii) ⇒ a+b = 13k ⋯ (iii) Adding (i),(ii),(iii) and dividing by 2 ⇒ a+b+c = 18k ∴ a=7k ...

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Byju's Answer

Standard XII

Mathematics

Properties of Argument

In Δ ABC, b+c...

Question

In $Δ A B C, \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then $sin A : sin B : sin C =$

7 : 6 : 5

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19 : 25 : 7

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19 : 7 : 25

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7 : 25 : 19

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Solution

The correct option is A $7 : 6 : 5$
Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = k;$
$\Rightarrow b + c = 11 k \dots (i)$
$\Rightarrow c + a = 12 k \dots (i i)$
$\Rightarrow a + b = 13 k \dots (i i i)$
Adding $(i), (i i), (i i i)$ and dividing by $2$
$\Rightarrow a + b + c = 18 k$
$∴ a = 7 k, b = 6 k, c = 5 k$
$sin A : sin B : sin C = 7 : 6 : 5$

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Similar questions

Q. In

Δ A B C

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

, then

cos C

is equal to

Q. In a

Δ A B C, \frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13},

then

{tan}^{2} (\frac{A}{2}) =

Q. In any

Δ A B C

\frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}

, then prove that

\frac{cos A}{2} = \frac{cos B}{7} = \frac{cos C}{11}

$In in any Δ A B C, \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}, then prove that \frac{c o s A}{2} = \frac{c o s B}{7} = \frac{c o s C}{11} .$