How to Find Missing Sides and Angles of a Right Triangle

1) Find the answer: $x \ = \ ?$

$\color{red}{tan \ 45° \ = \ 1 \ = \ \frac{AB}{BC} \ = \ \frac{x}{7} \ ⇒ \ x \ = \ \frac{\sqrt{2}}{2} \times 7 \ = \ \frac{7\sqrt{2}}{2}}$

2) Find the answer: $x \ = \ ?$

$\color{red}{tan \ x \ = \ \frac{AB}{BC} \ = \ \frac{5}{10} \ = \ \frac{1}{2} \ ⇒ \ x \ ≈ \ 26.565°}$

3) Find the answer: $x \ = \ ?$

$\color{red}{tan \ 60° \ = \ \sqrt{3} \ = \ \frac{AB}{BC} \ = \ \frac{6}{x} \ ⇒ \ x \ = \ \frac{6}{\sqrt{3}} \ = \ 2\sqrt{3}}$

4) Find the answer: $x \ = \ ?$

$\color{red}{tan \ 60° \ = \ \sqrt{3} \ = \ \frac{BC}{AB} \ = \ \frac{x}{11} \ ⇒ \ x \ = \ \sqrt{3} \times 11 \ = \ 11\sqrt{3}}$

5) Find the answer: $x \ = \ ?$

$\color{red}{tan \ x \ = \ \frac{BC}{AB} \ = \ \frac{4\sqrt{3}}{4} \ = \ \sqrt{3} \ ⇒ \ x \ = \ 60°}$

6) Find the answer: $sin \ x \ = \ ? \ , \ cos \ x \ = \ ?$

$\color{red}{AC \ = \ \sqrt{3^2 \ + \ 4^2} \ = \ 5}$
$\color{red}{sin \ x \ = \ \frac{BC}{AC} \ = \ \frac{4}{5}}$
$\color{red}{cos \ x \ = \ \frac{AB}{AC} \ = \ \frac{3}{5}}$

7) Find the answer: $cos \ x \ = \ ? \ , \ tan \ x \ = \ ?$

$\color{red}{BC \ = \ \sqrt{10^2 \ - \ 8^2} \ = \ \sqrt{36} \ = \ 6}$
$\color{red}{cos \ x \ = \ \frac{BC}{AC} \ = \ \frac{6}{10} \ = \ \frac{3}{5}}$
$\color{red}{tan \ x \ = \ \frac{AB}{AC} \ = \ \frac{8}{6} \ = \ \frac{4}{3}}$

8) Find the answer: $cot \ x \ = \ ? \ , \ sin \ x \ = \ ?$

$\color{red}{AB \ = \ \sqrt{15^2 \ - \ 12^2} \ = \ \sqrt{81} \ = \ 9}$
$\color{red}{cot \ x \ = \ \frac{BC}{AB} \ = \ \frac{12}{9} \ = \ \frac{4}{3}}$
$\color{red}{sin \ x \ = \ \frac{AB}{AC} \ = \ \frac{9}{15} \ = \ \frac{3}{5}}$

9) Find the answer: $x \ = \ ?$

$\color{red}{cos \ 45° \ = \ \frac{\sqrt{2}}{2} \ = \ \frac{BC}{AC} \ = \ \frac{9}{x} \ ⇒ \ x \ = \ 9 \times \frac{2}{\sqrt{2}} \ = \ \frac{\sqrt{2}}{9}}$

10) Find the answer: $x \ = \ ?$

$\color{red}{cos \ 49° \ ≈ \ 0.656 \ = \ \frac{BC}{AC} \ = \ \frac{9}{x} \ ⇒ \ x \ = \ 9 \times \frac{1}{0.656} \ ≈ \ 13.719}$