In triangle ABC, a = 5, b = 9, and c = 6. Find angle A, angle B, and angle C.
a. 38.9°, 109.5°, 331.6°
b.31.6°, 109.5°, 38.9°
c. 109.5°, 31.6°, 38.9°
d. 331.6°, 38.9°, 109.5°
well, first off you know that since a<c<b,
A<C<B (law of sines)
That means (b)
thanks steve for your help.
To find the angles of a triangle when you know the lengths of its sides, you can use the Law of Cosines. The Law of Cosines states:
c^2 = a^2 + b^2 - 2ab*cos(C)
where 'a', 'b', and 'c' are the side lengths of the triangle, and 'C' is the angle opposite to side 'c'.
Let's use this formula to find angle A, angle B, and angle C.
Given: a = 5, b = 9, and c = 6
First, let's find angle C:
c^2 = a^2 + b^2 - 2ab*cos(C)
6^2 = 5^2 + 9^2 - 2 * 5 * 9 * cos(C)
36 = 25 + 81 - 90*cos(C)
36 = 106 - 90 * cos(C)
90 * cos(C) = 106 - 36
90 * cos(C) = 70
cos(C) = 70 / 90
cos(C) = 0.7778
Next, use the inverse cosine function (cos^-1) to find angle C:
C = cos^-1(0.7778)
C = 39.2° (approximately)
Now, let's find angle A:
a^2 = b^2 + c^2 - 2bc*cos(A)
5^2 = 9^2 + 6^2 - 2 * 9 * 6 * cos(A)
25 = 81 + 36 - 108 * cos(A)
25 = 117 - 108 * cos(A)
108 * cos(A) = 117 - 25
108 * cos(A) = 92
cos(A) = 92 / 108
cos(A) = 0.8519
Again, use the inverse cosine function to find angle A:
A = cos^-1(0.8519)
A = 31.6° (approximately)
Finally, to find angle B, use the fact that the sum of the angles in a triangle is always 180°:
B = 180° - A - C
B = 180° - 31.6° - 39.2°
B = 109.2° (approximately)
Therefore, the correct answer is:
b. 31.6°, 109.2°, 39.2°