Answer:
(a) see below
(b) 36°
Explanation:
You want the angle between sides 12 mm and 23 mm in a triangle whose third side is 15 mm.
a) Angle
The cosine rule is given in the problem statement. Solving it for the angle, we have ...
2bc·cos(θ) +a² = b² +c² . . . . . . . . add 2bc·cos(θ)
2bc·cos(θ) = b² +c² -a² . . . . . . . . . subtract a²
cos(θ) = (b² +c² -a²)/(2bc) . . . . . . . divide by 2bc
b) Application
So, the angle is ...
θ = arccos((b² +c² -a²)/(2bc))
θ = arccos((12² +23² -15²)/(2·12·23))
θ ≈ 36°
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Additional comment
The calculator is in degrees mode.
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