For positive acute angles A and B, it is known that sinA= 8/17 and tanB= 40/9 ​ . Find the value of sin(A+B) in simplest form.

Answer 1

sinA = 8/17 tanB= 40/9

cosA = 15/17 sinB = 40/41, cosB = 9/41

sin(A+B) = sinAcosB+ sinBcosA= 72/17(41)+ 600/17(41)

= 672/697= about 0.9641

A = about 28.0725

B = about 77.3196

A + B = about 105.3921

sin105.3921 = about 0.9641

Answer 2

Draw a right triangle with acute angle A. Side opposite A has length 8 and hypotenuse has length 17. Use the Pythagorean Theorem to find the length of the third side. Draw another right triangle with acute angle B. Length of side opposite B is 40 and length of adjacent side to B is 9. Use the Pythagorean Theorem to find the length of the hypotenuse.

Then:

sinA = 8/17 and cosA = 15/17

tanB = 40/9 sinB = 40/41 and cosB = 9/41

sin(A+B) = sinAcosB + cosAsinB = (8/17)(9/41) + (15/17)(40/41) Simplify!