Question

find the sine, cosine, and tangent of ∠e.

write your answer in simplified, rationalized form. do not round.

sin (e) =
cos (e) =
tan (e) =

Explanation:

Step1: Find side $CE$ via Pythagoras

$$CE = \sqrt{(7\sqrt{2})^2 - (\sqrt{46})^2}$$
$$CE = \sqrt{98 - 46} = \sqrt{52} = 2\sqrt{13}$$

Step2: Define sides for $\angle E$

Opposite: $\sqrt{46}$, Adjacent: $2\sqrt{13}$, Hypotenuse: $7\sqrt{2}$

Step3: Calculate $\sin(E)$

$$\sin(E) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\sqrt{46}}{7\sqrt{2}} = \frac{\sqrt{46}\cdot\sqrt{2}}{7\cdot2} = \frac{\sqrt{92}}{14} = \frac{2\sqrt{23}}{14} = \frac{\sqrt{23}}{7}$$

Step4: Calculate $\cos(E)$

$$\cos(E) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2\sqrt{13}}{7\sqrt{2}} = \frac{2\sqrt{13}\cdot\sqrt{2}}{7\cdot2} = \frac{\sqrt{26}}{7}$$

Step5: Calculate $\tan(E)$

$$\tan(E) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{46}}{2\sqrt{13}} = \frac{\sqrt{46}\cdot\sqrt{13}}{2\cdot13} = \frac{\sqrt{598}}{26}$$

Answer:

$\sin(E) = \frac{\sqrt{23}}{7}$
$\cos(E) = \frac{\sqrt{26}}{7}$
$\tan(E) = \frac{\sqrt{598}}{26}$