The sum of angles of any triangle is 180 degrees
From the given figure
Since mSince m
Then subtract them from 180 degrees to find m
![\begin{gathered} m\angle B=180-90-53 \\ m\angle B=\mathring{37} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yk167zipp2pr646qzpyvr78e827nhzoqus.png)
The measure of angle B is 37 degrees
By using the trigonometry ratios we can find the lengths of AB and BC
Since the opposite side of Since the adjacent side of
We can use the tan C to find AB
![\begin{gathered} \tan C=\frac{opp.}{\text{adj.}} \\ \tan C=(AB)/(AC) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qubvq6qaab2spb96vn0iqa1us14nf81gjp.png)
Since Since AC = 13
![\tan 53=(AB)/(13)](https://img.qammunity.org/2023/formulas/mathematics/college/1ydmsc800o9i36y7xttdog5vaj1ovydulu.png)
By using cross multiplication
![\begin{gathered} AB=13*\tan 53 \\ AB=17.25158268 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9zr0dopu0fa8h9uk56aq8rkifp6debai0r.png)
Round it to 3 decimal places, then
AB = 17.252
Since BC is the hypotenuse of the triangle, then
We can use the cos ratio to find it
Since cos = adjacent/hypotenuse, then
![\begin{gathered} \cos C=(AC)/(BC) \\ \cos 53=(13)/(BC) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e7f4zcvk2fyo3r7tsnppzb87xcx0jd6bcd.png)
By using the cross multiplication
![BC*\cos 53=13](https://img.qammunity.org/2023/formulas/mathematics/college/oi54gnuk4euqibw05a7i06l0ist6prxa8j.png)
Divide both sides by cos 53
![\begin{gathered} BC=(13)/(\cos 53) \\ BC=21.60132183 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uwzng2r0f6lf8wlkfm11v4yzo2lyghp9gg.png)
Round it to 3 decimal places, then
BC = 21.601