Question

Solve the triangle.

A = 33°, a = 19, b = 14

Can this solved how ?

Answer 1

Answer on Question #42400 – Math – Geometry

Solve the triangle.

$$A = 33{}^\circ, a = 19, b = 14$$

Can this solved how?

Solution:

$$A = 33{}^\circ, a = 19, b = 14$$

Solving the triangle means to finding missing sides and angles (side c, angles C and B) Law of Sines (the Sine Rule):

$$\begin{array}{l} \frac{a}{\sin A} = \frac{b}{\sin B} \\ \sin B = \frac{b}{a} \sin A = \frac{14}{19} \cdot \sin 33{}^\circ \Rightarrow B = \arcsin \left(\frac{b}{a} \sin A\right) = \\ = \arcsin \left(\frac{14}{19} \cdot \sin 33{}^\circ\right) = 24{}^\circ \\ \end{array}$$ $$\text{or } B = 180{}^\circ - \arcsin \left(\frac{14}{19} \cdot \sin 33{}^\circ\right) = 156{}^\circ$$

Thus, we must consider two cases: $B = 24{}^\circ$ and $B = 156{}^\circ$

$$\#1 \ (B = 24{}^\circ)$$

The angles always add to $180{}^\circ$: when you know two angles you can find the third:

$$A + B + C = 180{}^\circ$$ $$C = 180{}^\circ - A - B = 180{}^\circ - 33{}^\circ - 24{}^\circ = 123{}^\circ$$

To find side c we can use law of sines again, but with side c and angle C:

$$c = \frac{\sin C}{\sin A} a = 19 \cdot \frac{\sin(123{}^\circ)}{\sin(33{}^\circ)} = 29$$ $$\#2 \ (B = 156{}^\circ)$$

The angles always add to $180{}^\circ$: when you know two angles you can find the third:

$$A + B + C = 180{}^\circ$$ $$C = 180{}^\circ - A - B = 180{}^\circ - 33{}^\circ - 156{}^\circ = -9{}^\circ$$

Angle can not be negative, so the second case is not possible.

Answer: $B = 24{}^{\circ}, C = 123{}^{\circ}, c = 29.$

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