In triangle abc, ab=5 bc=9 and ac=8
Choose the angle measures from greatest to least
The more standard way to state your question would have been..
In triangle ABC , a=9, b=8, and c=5
that is, let A, B, and C be the vertices, and a, b, and c be the sides opposite those vertices.
I usually find the largest angle first, using the cosine law. In this case angle A
9^2 = 5^2 + 8^2 - 2(5)(8)cos A
cos A = (25 + 64 - 81)/(80 = 8/80 = 1/10
angle A = 84.26°
Now use the Sine Law to find a second angle, the third is then easy.
AC=5 BC=9 AC=8
To find the angle measures in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that for any triangle ABC with sides a, b, and c, and angle C opposite side c, the law is defined as follows:
c^2 = a^2 + b^2 - 2ab*cos(C)
First, we need to find the value of angle A. Using the Law of Cosines,
a^2 = b^2 + c^2 - 2bc*cos(A)
5^2 = 9^2 + 8^2 - 2(9)(8)*cos(A)
25 = 81 + 64 - 144cos(A)
25 = 145 - 144cos(A)
-120 = -144cos(A)
cos(A) = -120 / -144
cos(A) = 0.8333
To find the value of angle A, we need to find the inverse cosine of 0.8333:
A = cos^(-1)(0.8333)
A ≈ 33.75 degrees
Similarly, we can find the value of angle B:
b^2 = a^2 + c^2 - 2ac*cos(B)
9^2 = 5^2 + 8^2 - 2(5)(8)*cos(B)
81 = 25 + 64 - 80cos(B)
81 = 89 - 80cos(B)
-8 = -80cos(B)
cos(B) = -8 / -80
cos(B) = 0.1
B = cos^(-1)(0.1)
B ≈ 84.26 degrees
Now, we have the angle measures:
Angle A ≈ 33.75 degrees
Angle B ≈ 84.26 degrees
Angle C = 180 - (33.75 + 84.26) degrees
Angle C ≈ 62.99 degrees
From greatest to least, the angle measures in triangle ABC are:
Angle B ≈ 84.26 degrees
Angle C ≈ 62.99 degrees
Angle A ≈ 33.75 degrees