# How do you find the missing sides and angles in the right triangle, where a is the side across from angle A, b across from B, and c across from the right angle given a = 28, c = 42?

Nov 17, 2017

See below.

#### Explanation:

First we need to find side b

Pythagoras Theorem states:

${a}^{2} + {b}^{2} = {c}^{2}$

So:

${b}^{2} = {c}^{2} - {a}^{2}$

$\therefore$

${b}^{2} = {\left(42\right)}^{2} - {\left(28\right)}^{2} = 1764 - 784 = 980 \implies b = \sqrt{980} = 14 \sqrt{5}$

There are a few different ways of finding the remaining angles, this is just one of them.

$\tan A = \frac{a}{b}$

$\therefore$

$\tan A = \frac{28}{14 \sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$

$A = {\tan}^{-} 1 \left(\tan A\right) = {\tan}^{-} 1 \left(\frac{2 \sqrt{5}}{5}\right) = {41.81}^{o}$

Angle B

The sum of the angles in a triangle add up to ${180}^{o}$

$\therefore$

${180}^{o} - \left({41.81}^{o} + {90}^{o}\right) = {48.19}^{o}$

Full solution:

$A = {41.81}^{o}$

$B = {48.19}^{o}$

$C = {90}^{o}$

$a = 28$

$b = 14 \sqrt{5} \approx 31.3$

$c = 42$