Question

1 sin a l \frac{14}{13} e \frac{5}{13}
2 cos a \frac{5}{12} v \frac{12}{5}
3 tan a

4 sin b p \frac{13}{5} y \frac{5}{13}
5 cos b
6 tan b w \frac{12}{13} e \frac{12}{5}

7 sin a w \frac{sqrt{3}}{2} \frac{1}{2}
8 cos a
9 tan a i \frac{1}{sqrt{3}} t 2

10 sin b e \sqrt{3} \frac{1}{2}
11 cos b
12 tan b a \frac{sqrt{3}}{2} i \frac{1}{sqrt{3}}

Explanation:

Step1: Identify triangle 1 sides

Right triangle: $BC=5$, $AC=12$, hypotenuse $AB=\sqrt{5^2+12^2}=13$

Step2: Solve for $\sin A$

$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{5}{13}$

Step3: Solve for $\cos A$

$\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}$

Step4: Solve for $\tan A$

$\tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AC}=\frac{5}{12}$

Step5: Solve for $\sin B$

$\sin B=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}$

Step6: Solve for $\cos B$

$\cos B=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{5}{13}$

Step7: Solve for $\tan B$

$\tan B=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{BC}=\frac{12}{5}$

Step8: Identify triangle 2 sides

Right triangle: $BC=1$, $AC=\sqrt{3}$, hypotenuse $AB=\sqrt{1^2+(\sqrt{3})^2}=2$

Step9: Solve for $\sin A$

$\sin A=\frac{BC}{AB}=\frac{1}{2}$

Step10: Solve for $\cos A$

$\cos A=\frac{AC}{AB}=\frac{\sqrt{3}}{2}$

Step11: Solve for $\tan A$

$\tan A=\frac{BC}{AC}=\frac{1}{\sqrt{3}}$

Step12: Solve for $\sin B$

$\sin B=\frac{AC}{AB}=\frac{\sqrt{3}}{2}$

Step13: Solve for $\cos B$

$\cos B=\frac{BC}{AB}=\frac{1}{2}$

Step14: Solve for $\tan B$

$\tan B=\frac{AC}{BC}=\sqrt{3}$

Answer:

1. $\boldsymbol{E.\ \frac{5}{13}}$
2. $\boldsymbol{L.\ \frac{12}{13}}$
3. $\boldsymbol{\frac{5}{12}}$
4. $\boldsymbol{W.\ \frac{12}{13}}$
5. $\boldsymbol{Y.\ \frac{5}{13}}$
6. $\boldsymbol{E.\ \frac{12}{5}}$
7. $\boldsymbol{\frac{1}{2}}$
8. $\boldsymbol{W.\ \frac{\sqrt{3}}{2}}$
9. $\boldsymbol{I.\ \frac{1}{\sqrt{3}}}$
10. $\boldsymbol{A.\ \frac{\sqrt{3}}{2}}$
11. $\boldsymbol{\frac{1}{2}}$
12. $\boldsymbol{E.\ \sqrt{3}}$