Suppose 𝜃∘ and 𝜙 are in the first quadrant and that sin(𝜃∘)=3/17 and cos(𝜙)=2/13. Then, determine the following

Question

Determine sin(𝜋/2−𝜙) and cos(90∘−𝜃∘). I tried to figure this out but the answer will not let me put in any kind of decimal points so I am not sure how to move forward.

Answer 1

sinA = 3/17, use an inverse trig function on a calculator

A = about 10.164 degrees

90- 10.164 = 79.836

sin79.836 = 0.984

or if sinA = 3/17 = opposite side over hypotenuse, then cosA = adjacent side over hypotenuse

adjacent side squared = 17^2 - 3^2 = 289-9 = 280

adjacent side = sqr280 = 2sqr70

cosA = sin(90-A) = 2sqr70/17 = about 0.9843

cosB = 2/13 sinB = sqr(13^-2^2)/13 = sqr(169-4)/13 = sqr165/13

cos(90-B) = sinB = sqr164/13 = about 0.9881

major point of the problem is cos(90-theta) = sin(theta) and sin(90-theta) = cos(theta)