# How do you find the measure of each of the angles of a triangle given the measurements of the sides are 30, 35, 45?

Jan 3, 2018

Angle opposite to 30 $\approx 41.75$ degrees
Angle opposite to 35 $\approx 59.98$ degrees
Angle opposite to 45 $\approx 87.27$ degrees

#### Explanation:

Use the law of cosines: ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(C\right)$

However, we require angle measures, therefore, we can put the entire equation in terms of $C$:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(C\right)$

$\implies {c}^{2} - {a}^{2} - {b}^{2} = - 2 a b \cdot \cos \left(C\right)$

$\implies \frac{{c}^{2} - {a}^{2} - {b}^{2}}{- 2 a b} = \cos \left(C\right)$

$\implies \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b} = \cos \left(C\right)$

$\implies C = \arccos \left(\frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}\right)$

Angle opposite to 30:

$\arccos \left(\frac{{35}^{2} + {45}^{2} - {30}^{2}}{2 \cdot 35 \cdot 45}\right)$

$= \arccos \left(\frac{2350}{3150}\right)$

$= \arccos \left(\frac{47}{63}\right) \approx \ast 41.75 \ast$

Angle opposite to 35:

$\arccos \left(\frac{{30}^{2} + {45}^{2} - {35}^{2}}{2 \cdot 30 \cdot 45}\right)$

$= \arccos \left(\frac{1700}{2700}\right)$

$= \arccos \left(\frac{17}{27}\right) \approx \ast 59.98 \ast$

Angle opposite to 45:

$\arccos \left(\frac{{30}^{2} + {35}^{2} - {45}^{2}}{2 \cdot 30 \cdot 35}\right)$

$= \arccos \left(\frac{100}{2100}\right)$

$= \arccos \left(\frac{1}{21}\right) \approx \ast 87.27 \ast$