The sines of the two acute angles of the triangle are 7/25 and 4/5, respectively. Find the cosine of the third corner of the triangle.

Let us denote the angles of a triangle whose sines are 7/25 and 4/5, respectively, by α and β, and the third angle of this triangle by γ.

Since the first two corners of the triangle are sharp, their cosines are positive.

Find cosα and cosβ:

cosα = √ (1 – sin ^ 2α) = √ (1 – (7/25) ^ 2) = √ (1 – 49/625) = √ (576/625) = 24/25;

cosβ = √ (1 – sin ^ 2β) = √ (1 – (4/5) ^ 2) = √ (1 – 16/25) = √ (9/25) = 3/5.

Find cos (α + β):

cos (α + β) = cos (α) cos (β) – sin (α) sin (β) = (24/25) * (3/5) – (7/25) * (4/5) = 44 / 125.

Find cosγ:

cosγ = cos (180 ° – (α + β)) = -cos (α + β) = -44/125.

Answer: -44/125.