Question

Use the Law of Cosines, a2=b2+c2−2bc⋅cosA , to find the value of angle A in a right triangle, where a=9 , b=40 , and c=41 .

Answer 1

ah, but it says it is a right triangle.
Yes in fact it is. 41^2 = 40^2 + 9^2 or 1681 = 1600 + 81
so without further ado I think cos A = 40/41 = 0.97561
and agreed A = 12.68 degrees

Answer 2

In this case, we have a=9, b=40, and c=41. To find angle A, we can plug these values into the Law of Cosines formula:

a^2 = b^2 + c^2 - 2bc cos(A)
9^2 = 40^2 + 41^2 - 2(40)(41)cos(A)
81 = 1600 + 1681 - 3280cos(A)
81 = 3281 - 3280cos(A)
3280cos(A) = 3200
cos(A) = 3200/3280
cos(A) = 0.97561

Now we need to find the angle whose cosine is approximately 0.97561. Using a calculator, we can calculate the inverse cosine of 0.97561 to find angle A:

A ≈ cos^(-1)(0.97561)
A ≈ 12.69 degrees

Therefore, in the given right triangle, angle A measures approximately 12.69 degrees.

Answer 3

Yes, that is correct! Since the triangle is a right triangle with sides satisfying the Pythagorean theorem, angle A must be the acute angle opposite the side of length 9. Therefore, angle A measures approximately 12.68 degrees. Thank you for pointing that out!