Question

In a ΔABC, A is obtuse. If sin A = $\frac{3}{5}$ and sin B = $\frac{5}{13}$, then show that sin C = $\frac{16}{65}$.

Answer 1

we know , if ABC is a triangle

then, A + B + C = 180°

A + B = 180° - C

sin(A + B) = sin(180° - C)

sinA.cosB + cosA.sinB = sinC

given, sinA = 3/5 = p/h

so, p = 3 and h = 5

then, b = √(h² - p²) = √(5² - 3²) = ± 4

cosA = b/h = ±4/5 but A is obtuse angle (e.g., 0 < A < 180°)
so, cosA = -4/5

Similarly, sinB = 5/13

then, cosB = 12/13 [ because B is acute angle]

now, sinA.cosB + cosA.sinB = sinC

3/5 × 12/13 + (-4/5) × 5/13 = sinC

36/65 - 20/65 = sinC

16/65 = sinC [hence proved]

Answer 2

HELLO DEAR,



GIVEN:- sinA = 3/5 , cosA = √{(25 - 9)/25} = 4/5[but A is obtuse so, cosA = -4/5]

sinB = 5/13 , cosB = √{(169 - 25)/169} = 12/13


and, in any ∆ABC , sum of angles of a triangle are 180°

e.g., A + B + C = 180°


A + B = (180 - C)



now,
sin(A + B) = sinA.cosB + cosA.sinB

=> sin(180 - C) = (3/5)(12/13) + (-4/5)(5/13)

=> sinC = 36/65 - 20/65

=> sinC = 16/65


I HOPE IT'S HELP YOU DEAR,
THANKS