# How do you solve the right triangle ABC given a=3, A=26?

In right $\setminus \triangle A B C$, we have $\setminus \angle A = {26}^{\setminus} \circ$ & $a = 3$ then other acute angle $\setminus \angle C$ is given as

$\setminus \angle C = {180}^{\setminus} \circ - \setminus \angle B - \setminus \angle A$

$= {180}^{\setminus} \circ - {90}^{\setminus} \circ - {26}^{\setminus} \circ$

$= {64}^{\setminus} \circ$

Now, by applying Sine rule in right $\setminus \triangle A B C$ as follows

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

$\setminus \frac{3}{\setminus \sin {26}^{\setminus} \circ} = \setminus \frac{b}{\setminus \sin {64}^{\setminus} \circ} = \setminus \frac{c}{\setminus \sin {90}^{\setminus} \circ}$

Consider,

$\setminus \frac{3}{\setminus \sin {26}^{\setminus} \circ} = \setminus \frac{b}{\setminus \sin {64}^{\setminus} \circ}$

$b = \setminus \frac{3 \setminus \sin {64}^{\setminus} \circ}{\setminus \sin {26}^{\setminus} \circ}$

$= 6.151$

Consider,

$\setminus \frac{3}{\setminus \sin {26}^{\setminus} \circ} = \setminus \frac{c}{\setminus \sin {90}^{\setminus} \circ}$

$c = \setminus \frac{3 \setminus \sin {90}^{\setminus} \circ}{\setminus \sin {26}^{\setminus} \circ}$

$= 6.843$