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In a right triangle, Angle A and B are acute. If tan B= 2, what is cos B? Round to three decimal places.


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oladayoademosu

The value of cosB ≅ 0.447 to three decimal places.

Since angle A and angle B are acute and tanB = 2, we find cosB from the trigonometric identity

tan²B + 1 = sec²B

Find SecB

So, substituting the value of tanB into the equation, we have

tan²B + 1 = sec²B

2² + 1 = sec²B

4 + 1 = sec²B

5 = sec²B

Taking square root of both sides, we have

√sec²B = ±√5

secB = ±√5

Find CosB

Since secB = 1/cosB, we have that

1/cosB = ±√5

⇒cosB = ±1/√5

Since B is acute, cosB will be positive.

So, cosB = 1/√5

cosB = 1/2.2361

cosB = 0.4472

cosB ≅ 0.447 to three decimal places.

So, the value of cosB ≅ 0.447 to three decimal places.

Learn more about cosine here:

https://brainly.com/question/10417664

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