Question

In a triangle ABC, if a=18 , b=24, c=30 find :- sinA, sinB ,sinC ?

Answer 1

"Given lengths represent 'Pythagorean triplet' (18²+24²)=(30)²

a = 18, b = 24, c = 30 gives C = 90 degrees as a^2 + b^2 = c^2 then

cos A = b/c = 24/30 = 4/5 ANSWER

verify trigonometrically

cos A = (b^2 + c^2 -- a^2) / 2bc = (24^2 + 30^2 -- 18^2) / 2*24*30 = 1152/1440 = 4/5

cos B = a/c = 18/30 = 3/5 ANSWER

cos C = cos (90) = 0 ANSWER

sin A = a/c = 18/30 = 3/5 ANSWER

sin B = b/c = 24/30 = 4/5 ANSWER

sin C = sin (90) = 1 ANSWER

"

Answer 2

Answer:

Step-by-step explanation:

Given lengths represent 'Pythagorean triplet' (18²+24²)=(30)²

a = 18, b = 24, c = 30 gives C = 90 degrees as a^2 + b^2 = c^2 then

cos A = b/c = 24/30 = 4/5 ANSWER

verify trigonometrically

cos A = (b^2 + c^2 -- a^2) / 2bc = (24^2 + 30^2 -- 18^2) / 2*24*30 = 1152/1440 = 4/5

cos B = a/c = 18/30 = 3/5 ANSWER

cos C = cos (90) = 0 ANSWER

sin A = a/c = 18/30 = 3/5 ANSWER

sin B = b/c = 24/30 = 4/5 ANSWER

sin C = sin (90) = 1 ANSWER