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1 year ago

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.

Mathematics

2 answers:

Westkost [7]1 year ago

8 0

Answer:

cos <ABC=0

cos <BAC=3/5

cos<ACB=4/5

GIVE 5 STARS IF YOU DO RSM!!!

pickupchik [31]1 year ago

7 0

Answer:

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

Step-by-step 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

<h3>Angle A</h3>

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

<h3>Angle B</h3>

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

cos(B) = 0

<h3>Angle C</h3>

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

cos(C) = CB/AC

cos(C) = 4/5

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