In a triangle ABC , A is obtuse , sinA=3/5 , sinB=5/13 then sinC=?

Open in App

Solution

sinA = 3/5 so cosA = - 4/5 (since A obtuse); similarly, using a simple right-angled triangle if cosB = 12/13, then sinB = 5/13. C = 180 - (A + B) so sinC = sin[180 - (A + B)] = sin(A + B) (4 quadrants) so sinC = sinAcosB + cosAsinB = (3/5)(12/13) + (-4/5)(5/13) = 36/65 - 20/65 = 16/65

Suggest Corrections

Similar questions

Δ l e A B C

A

S i n A = 3 / 5, S i n B = 5 / 13

S i n C =

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

sin A = sin B

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

△ A B C, sin A = \frac{4}{5}

sin B = \frac{12}{13}

sin C =

A, B, C

△ A B C

(s i n A + s i n B) (s i n B + s i n C) (s i n C + s i n A)

A B C,

sin A, sin B, sin C

Join BYJU'S Learning Program