Question

test
solve each triangle. round your answers to the nearest tenth.

1. (mangle a = 133^{circ},mangle b = 22^{circ},a = 45)
2. (c = 30m,mangle a=123^{circ},b = 21m)

Explanation:

Step1: Find angle C in the first triangle

The sum of angles in a triangle is 180°. So, $m\angle C=180^{\circ}-m\angle A - m\angle B$. Substituting $m\angle A = 133^{\circ}$ and $m\angle B=22^{\circ}$, we get $m\angle C=180^{\circ}-133^{\circ}-22^{\circ}=25^{\circ}$.

Step2: Use the Law of Sines to find side b in the first triangle

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $a = 45$, $m\angle A=133^{\circ}$, and $m\angle B = 22^{\circ}$. So, $b=\frac{a\sin B}{\sin A}=\frac{45\times\sin(22^{\circ})}{\sin(133^{\circ})}\approx22.3$.

Step3: Use the Law of Sines to find side c in the first triangle

Using the Law of Sines $\frac{a}{\sin A}=\frac{c}{\sin C}$, with $a = 45$, $m\angle A=133^{\circ}$, and $m\angle C = 25^{\circ}$. Then $c=\frac{a\sin C}{\sin A}=\frac{45\times\sin(25^{\circ})}{\sin(133^{\circ})}\approx25.8$.

For the second - triangle:

Step4: Find angle C using the Law of Cosines

First, use the Law of Cosines $c^{2}=a^{2}+b^{2}-2ab\cos C$. We are given $c = 30$, $a = 21$, and $m\angle A=123^{\circ}$. First, find angle C.
$c^{2}=a^{2}+b^{2}-2ab\cos C$ can be rewritten as $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$. Substituting $a = 21$, $b$ (not given in a way to use this directly, let's use the Law of Sines).
Using the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. First, find $m\angle B$.
$m\angle B=180^{\circ}-m\angle A - m\angle C$.
Using the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$, we know $a = 21$, $m\angle A = 123^{\circ}$, $c = 30$.
$\sin B=\frac{b\sin A}{a}$.
First, find $m\angle C$:
$\frac{a}{\sin A}=\frac{c}{\sin C}$, so $\sin C=\frac{c\sin A}{a}=\frac{30\times\sin(123^{\circ})}{21}\approx1.19$ (This is wrong. Let's start over using the Law of Cosines correctly)
Using the Law of Cosines $c^{2}=a^{2}+b^{2}-2ab\cos C$
$900 = 441+ b^{2}-42b\cos(123^{\circ})$
$900=441 + b^{2}+ 22.3b$
$b^{2}+22.3b - 459 = 0$
Using the quadratic formula $b=\frac{-22.3\pm\sqrt{22.3^{2}-4\times(- 459)}}{2}=\frac{-22.3\pm\sqrt{497.29 + 1836}}{2}=\frac{-22.3\pm\sqrt{2333.29}}{2}=\frac{-22.3\pm48.3}{2}$
We take the positive root $b=\frac{-22.3 + 48.3}{2}=13$
Now, using the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$
$\sin B=\frac{b\sin A}{a}=\frac{13\times\sin(123^{\circ})}{21}\approx0.52$
$m\angle B\approx31.3^{\circ}$
$m\angle C=180^{\circ}-123^{\circ}-31.3^{\circ}=25.7^{\circ}$
Using the Law of Sines to find $b$:
$\frac{a}{\sin A}=\frac{b}{\sin B}$, $b=\frac{a\sin B}{\sin A}=\frac{21\times\sin(31.3^{\circ})}{\sin(123^{\circ})}\approx13.0$

Answer:

For the first triangle:
$m\angle C = 25^{\circ}$, $b\approx22.3$, $c\approx25.8$
For the second triangle:
$m\angle B\approx31.3^{\circ}$, $m\angle C\approx25.7^{\circ}$, $b\approx13.0$