Question

the vertices of ABC are points A(1,1), B(4,1), and C(4,5). find the cosines of the angles of the triangle.

Answer 1

cos <ABC=0

cos <BAC=3/5

cos<ACB=4/5

GIVE 5 STARS IF YOU DO RSM!!!

Answer 2

• cos(A) = 3/5
• cos(B) = 0
• cos(C) = 4/5

Explanation:

The mnemonic SOH CAH TOA reminds you of the relation between the cosine of an angle and the sides of the triangle.

Cos = Adjacent/Hypotenuse

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Angle A

In the given triangle, the hypotenuse is AC. The side adjacent to angle A is AB, so its cosine is ...

cos(A) = AB/AC

cos(A) = 3/5

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Angle B

The right angle in the triangle is angle B. The cosine of a right angle is 0.

cos(B) = 0

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Angle C

The side adjacent to angle C is CB, so its cosine is ...

cos(C) = CB/AC

cos(C) = 4/5