# How do you solve the right triangle ABC given b=3, B=26?

Oct 14, 2017

See below.

#### Explanation:

I am assuming $B = 26$ refers to the measurement of angle B in degrees.

Angle A = ${90}^{o} - {26}^{o} = {64}^{o}$

Angle B = ${26}^{o}$

Angle C = ${90}^{o}$

Side b = 3

Since we know all three angles and one side, we can use the Sine Rule to solve this:

$\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$

We will use $\sin \frac{A}{a} = \sin \frac{B}{b}$ , because we know angles A and B and we know b.

So:

$\sin \frac{64}{a} = \sin \frac{26}{3} \implies a = \frac{3 \sin \left(64\right)}{\sin} \left(26\right) = 6.151$ (3 .d.p)

By Pythagoras' Theorem:

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

${c}^{2} = {\left(\frac{3 \sin \left(64\right)}{\sin} \left(26\right)\right)}^{2} + {3}^{2} \implies c = \sqrt{{\left(\frac{3 \sin \left(64\right)}{\sin} \left(26\right)\right)}^{2} + {3}^{2}} = 6.844$ ( 3 .d.p)

So we have solved the right angled triangle:

$a = 6.151$ (3 .d.p)

$b = 3$

$c = 6.844$ ( 3 .d.p)

$A = {64}^{o}$

$B = {26}^{o}$

$C = {90}^{o}$