Angle Sum in Triangle Abc
In triangle $ ABC, $ $ a = 2, $ $ b = 3, $ and $ c = 4 $. Find $ 3 \angle A + 2 \angle B, $ in degrees.
Note: $ a $ is the side length opposite $ \angle A, $ etc.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
By the Law of Cosines,
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3^2 + 4^2 - 2^2}{2 \cdot 3 \cdot 4} = \frac{7}{8},\]and
\[\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{2^2 + 4^2 - 3^2}{2 \cdot 2 \cdot 4} = \frac{11}{16}.\]Then
\[\cos 3A = 4 \cos^3 A - 3 \cos A = 4 \left( \frac{7}{8} \right)^3 - 3 \cdot \frac{7}{8} = \frac{7}{128},\]and
\[\cos 2B = 2 \cos^2 B - 1 = 2 \left( \frac{11}{16} \right)^2 - 1 = -\frac{7}{128}.\]Since $ \cos 3A + \cos 2B = 0, $ either $ 3A + 2B $ is an odd multiple of $ 180^\circ, $ or $ 3A - 2B $ is an odd multiple of $ 180^\circ $.
Since $ \cos A $ and $ \cos B $ are positive, both $ A $ and $ B $ are acute. Furthermore, $ \cos 3A $ is positive, so $ 3A < 90^\circ $. Hence,
\[-180^\circ < 3A - 2B < 90^\circ,\]which means $ 3A - 2B $ cannot be an odd multiple of $ 180^\circ $. Therefore, $ 3A + 2B $ must be an odd multiple of $ 180^\circ $. But
\[3A + 2B < 90^\circ + 180^\circ = 270^\circ,\]which means $ 3A + 2B = \boxed{180^\circ} $.
Freeflows