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2 years ago

7

If 41 sin A = 40, find the value of cos A and tan A.

1 answer:

lorasvet [3.4K]2 years ago

5 0

Answer:

cos A= 9/41 and tan A is 40/9

Step-by-step explanation:

SOHCAHTOA.

Sine = opposite/hypotenuse

Cosine=adjacent/hypotenuse

Tangent=opposite/adjacent

Let's go back to your problem

41 sin A = 40

To get sin A by itself, divide both sides of the equation by 41. You get sin A = 40/41. As sine = opposite/hypotenuse, 40 will be the leg and 41 will be the hypotenuse. What is the value of the other leg?

Use the pythagorean theorem.

a^2 + b^2 = c^2

a^2 + 40^2 = 41^2

a^2 + 1600 = 1681

a^2 = 81

a=9

This is a 9-40-41 right triangle

Now, solve cos A

cos=adjacent/hypotenuse

cos=9/41

Finally, solve for tan A

tan=opposite/adjacent

tan=40/9

Hope this helps :)

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To proof

As given

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Step-by-step explanation:

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