How do you solve the triangle ABC given a=55, b=25, c=72?

Aug 24, 2016

hatA = 39.3°, hatB = 16.8° , hatC = 123.9°

Explanation:

"Solve a triangle" means find all the missing information - in this case, the size of each angle.

The lengths of 3 sides are given. We must use the Cosine Rule.

Find the biggest angle first - if it is an obtuse angle, the Cos rule will indicate this, but the Sin Rule will not,

Angle C is the biggest angle because it is opposite the longest side.

$C o s \hat{C} = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b} = \frac{{55}^{2} + {25}^{2} - {72}^{2}}{2 \times 55 \times 25}$

$C o s \hat{C} = - 0.5578 \text{ (the negative value means it is obtuse)}$

color(red)(hatC =123.9°)

Now use the SIn Rule.

$\frac{S \in \hat{A}}{a} = \frac{S \in \hat{C}}{c}$

$S \in \hat{A} = \frac{55 \times S \in 123.9}{72} = 0.6340$

color(red)(hatA = 39.3°)

We have 2 angles, use the angle sum of a triangle to find $\hat{B}$

hatB = 180° - 123.9° - 39.3°

color(red)(hatB = 16.8°)