In triangle ABC: angle C = 90 degrees, AB = 5, BC = 3. find the cosine of the outer angle at the vertex A

By the Pythagorean theorem, we determine the length of the leg AC.

AC ^ 2 = AB ^ 2 – BC ^ 2 = 25 – 9 = 16.

AC = 4 cm.

Then CosBAC = AC / AB = 4/5 = 0.8.

Since the ВAD angle is adjacent to the BAC angle, then CosBAD = -CosBAC = -0.8.

Second way.

Determine the sine of the angle BAC.

SinBAC = BC / AB = 3/5 = 0.6.

Then CosBAC = √ (1 – Sin2BAC) = √ (1 – 0.36) = 0.8.

CosBAD = -CosBAC = -0.8, since the corners are adjacent.

Answer: The cosine of the outer corner is -0.8.