There exists a triangle $A B C$ satisfying

t a n A + t a n B + t a n C = 0

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\frac{s i n A}{2} = \frac{s i n B}{3} = \frac{s i n C}{7}

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(a + b)^{2} = c^{2} +

ab and

\sqrt{2} (s i n A + c o s A) = \sqrt{3}

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a + b = c

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Solution

The correct option is C $(a + b)^{2} = c^{2} +$ ab and $\sqrt{2} (s i n A + c o s A) = \sqrt{3}$
$\to t a n A + t a n B + t a n C = t a n A t a n B t a n C$
$\to \frac{s i n A}{2} = \frac{s i n B}{3} = \frac{s i n C}{7}$
$∴ a = 2 k, b = 3 k, c = 7 k$
$a + b > c$ (which is not happening in-thing case)
$\to a + b > c$
$\to a^{2} + b^{2} + 2 a b = c^{2} + a b$
$\frac{a^{2} + b^{2} - c^{2}}{a b} = - 1$
$\frac{- 1}{2} = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
$∴ c o s C = \frac{- 1}{2}$
$c = \frac{2 π}{3}$
$∴ A + B = \frac{π}{3}$
$s i n 2 A + c o s 2 A + 2 s i n A c o s A = \frac{3}{2}$
$1 + s i n 2 A = \frac{3}{2}$
$s i n 2 A = \frac{1}{2}$
$2 A = \frac{π}{6}$
$A = \frac{π}{12}$

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Similar questions

t a n A + t a n B + t a n C = 0

\frac{s i n A}{2} = \frac{s i n B}{3} = \frac{s i n C}{7}

(a + b)^{2} = c^{2} + a b

\sqrt{2} (s i n A + c o s A) = \sqrt{3}

s i n A + s i n B = \frac{\sqrt{3} + 1}{2}

c o s A c o s B = \frac{\sqrt{3}}{4} = s i n A s i n B

A B C

{(a + b)}^{2} = c^{2} + a b

sin A + sin B + sin C = 1 + \frac{\sqrt{3}}{2}

(a + b)^{2} = c^{2} + a b

\sqrt{2} (s i n A + c o s A) = \sqrt{3} .

sin A + sin B + sin C = 1 + \frac{\sqrt{3}}{2}

sin A = sin B

{(a + b)}^{2} = c^{2} + a b

If cos(A-B) = 3/5 and tanA*tanB = 2 then

(a) cosA*cosB= 1/5 (c) cos(A+B)= -1/5

(b) sinA*sinB= 2/5 (d) sinA*sinB=4/5

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