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ΔABC is a triangle with sides measuring 5 cm, 7 cm, and 9cm, using the law of cosines

find ∠A, ∠B, and ∠C.

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The Law of Cosines states that for a triangle with sides a, b, and c and with angle C opposite side c, the relationship between the sides and the cosine of the angle is:

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

Since the triangle is a right triangle, we can use the Pythagorean theorem to find the value of the angle.

For ∠A:

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

9^2 = 5^2 + 7^2 - 257 * cos(A)

81 = 25 + 49 - 70 * cos(A)

81 = 74 - 70 * cos(A)

81 - 74 = -70 * cos(A)

7 = -70 * cos(A)

cos(A) = -7/70

A = arccos(-7/70)

For ∠B:

a^2 = b^2 + c^2 - 2bc * cos(B)

5^2 = 7^2 + 9^2 - 279 * cos(B)

25 = 49 + 81 - 98 * cos(B)

25 = 130 - 98 * cos(B)

130 - 25 = 98 * cos(B)

105 = 98 * cos(B)

cos(B) = 105/98

B = arccos(105/98)

For ∠C:

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

7^2 = 5^2 + 9^2 - 259 * cos(C)

49 = 25 + 81 - 90 * cos(C)

49 = 106 - 90 * cos(C)

106 - 49 = 90 * cos(C)

57 = 90 * cos(C)

cos(C) = 57/90

C = arccos(57/90)

Keep in mind that the values of the angle are in radians.

Please note that the triangle does not have to be a right triangle, the Law of Cosines works for any triangle.

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