In triangle ABC, the angle C is 90 °, cosA = 7: 25. Find the cosine of the outer corner at vertex B.

Let us write down the definition of the cosine for angle A.
Cos A = AC / AB = 7/25.
Let’s introduce the coefficient of proportionality k and get that AC = 7k, AB = 25k. Find the unknown BC leg.
ВС = √ (AB² – AC²) = √ (625k² – 49k²) = √576k² = 24k.
Let’s write down and find the cosine of the angle B.
Cos B = BC / AB = 24k / 25k = 24/25.
Outside angle at apex B is adjacent to angle B of triangle ABC. The cosine of the outer corner is equal to the cosine of the inner corner, taken with the opposite sign:
Cos (180 ° – B) = – cos B = – 24/25 = – 0.96.
Answer: cosine of the outer corner (- 0.96).