In triangle $A B C$ , If $tan (A / 2) = 5 / 6$ and $tan (B / 2) = 20 / 37$ , the sides a, b, and c are in

A.P

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G.P

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H.P

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none of these

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Solution

The correct option is D none of these
$tan (\frac{A}{2}) = \frac{Δ}{s (s - a)} = \frac{5}{6}$ (given) ......(i)
and $tan (\frac{B}{2}) = \frac{Δ}{s (s - b)} = \frac{20}{37}$ (given) ......(ii)
Divide (i) by (ii)
$\Rightarrow \frac{s - b}{s - a} = \frac{37}{24}$
$\Rightarrow 24 s - 24 b = 37 s - 37 a$

$\Rightarrow 13 s = 37 a - 24 b$

$\Rightarrow 13 (a + b + c) = 74 a - 48 b$
$\Rightarrow 61 b + 13 c = 61 a$

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Similar questions

Δ A B C

tan (A / 2) = \frac{5}{6}

tan (B / 2) = \frac{20}{37}

a, b

c

In a $△ A B C$ , if $t a n \frac{A}{2} = \frac{5}{6} a n d t a n \frac{B}{2} = \frac{20}{37}$ ,then

tan \frac{A}{2} = \frac{5}{6}, tan \frac{B}{2} = \frac{20}{37}

tan \frac{C}{2} = \frac{2}{5}

tan \frac{C}{2} = cot (\frac{A + B}{2})

Δ

A B C

A, B, C

a, b, c

tan \frac{A}{2} = \frac{5}{6}

tan \frac{B}{2} = \frac{20}{37}

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