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Consider a triangle ABC like the one below. Suppose that a=19, b=45, and c=41. (The figure is not drawn to scale.) solve the triangle. Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. If there is more than one solution, use the button labeled “or”.

Consider a triangle ABC like the one below. Suppose that a=19, b=45, and c=41. (The-example-1

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Answer:

A ≈ 25.6°, B ≈ 52.3°, C ≈ 68.8° or A

Explanation:

To solve the triangle ABC, we can use the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of the angles opposite those sides. The law states that:

c^2 = a^2 + b^2 - 2ab cos(C)

where c is the length of the side opposite angle C, a is the length of the side opposite angle A, and b is the length of the side opposite angle B.

Using the given values of a, b, and c, we can substitute into the formula and solve for cos(C):

41^2 = 19^2 + 45^2 - 2(19)(45) cos(C)

1681 = 361 + 2025 - 1710 cos(C)

cos(C) = (361 + 2025 - 1681) / (21945)

cos(C) = 0.37103

Now we can use the inverse cosine function to find the measure of angle C:

C = cos^-1(0.37103)

C ≈ 68.8°

Next, we can use the Law of Sines to find the measures of angles A and B:

sin(A)/a = sin(C)/c

sin(A)/19 = sin(68.8°)/41

sin(A) ≈ 0.4250

A ≈ 25.6°

Similarly, we can find the measure of angle B:

sin(B)/b = sin(C)/c

sin(B)/45 = sin(68.8°)/41

sin(B) ≈ 0.7873

B ≈ 52.3°

Therefore, the solution for the triangle ABC is:

A ≈ 25.6°, B ≈ 52.3°, C ≈ 68.8° or A

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