Question

find the sine, cosine, and tangent of $\angle g$.

(image of right triangle efg with right angle at f, ef = 7, fg = $\sqrt{22}$)

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

$\sin(g) = \square$
$\cos(g) = \square$
$\tan(g) = \square$

Explanation:

Step1: Find the hypotenuse \( EG \)

In right triangle \( EFG \), by Pythagorean theorem \( EG^2 = EF^2 + FG^2 \). Given \( EF = 7 \), \( FG=\sqrt{22} \), so \( EG^2=7^2 + (\sqrt{22})^2=49 + 22 = 71 \), then \( EG=\sqrt{71} \).

Step2: Calculate \( \sin(G) \)

\( \sin(G)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{EF}{EG}=\frac{7}{\sqrt{71}} \). Rationalize the denominator: \( \frac{7\sqrt{71}}{71} \).

Step3: Calculate \( \cos(G) \)

\( \cos(G)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{FG}{EG}=\frac{\sqrt{22}}{\sqrt{71}} \). Rationalize the denominator: \( \frac{\sqrt{1562}}{71}=\frac{11\sqrt{14}}{71} \)? Wait, no, \( \sqrt{22}\times\sqrt{71}=\sqrt{22\times71}=\sqrt{1562} \)? Wait, no, correct rationalization: \( \frac{\sqrt{22}}{\sqrt{71}}=\frac{\sqrt{22}\times\sqrt{71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, 22 and 71 are coprime, so \( \cos(G)=\frac{\sqrt{22\times71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, original: \( FG = \sqrt{22} \), \( EG=\sqrt{71} \), so \( \cos(G)=\frac{\sqrt{22}}{\sqrt{71}}=\frac{\sqrt{22}\times\sqrt{71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, 2271=1562, but maybe I made a mistake. Wait, Pythagorean theorem: \( EF = 7 \), \( FG=\sqrt{22} \), so \( EG=\sqrt{7^2 + (\sqrt{22})^2}=\sqrt{49 + 22}=\sqrt{71} \), correct. Then \( \cos(G)=\frac{FG}{EG}=\frac{\sqrt{22}}{\sqrt{71}}=\frac{\sqrt{22}\times\sqrt{71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, 22 and 71 are both primes? 22 is 211, 71 is prime. So \( \sqrt{22}\times\sqrt{71}=\sqrt{22\times71}=\sqrt{1562} \), but 1562=21171, so no square factors. So \( \cos(G)=\frac{\sqrt{1562}}{71} \)? Wait, no, maybe I messed up. Wait, adjacent to angle G is FG, opposite is EF, hypotenuse EG. So \( \sin(G)=\frac{EF}{EG}=\frac{7}{\sqrt{71}}=\frac{7\sqrt{71}}{71} \), \( \cos(G)=\frac{FG}{EG}=\frac{\sqrt{22}}{\sqrt{71}}=\frac{\sqrt{22}\times\sqrt{71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, 22*71=1562, but maybe I should write it as \( \frac{\sqrt{1562}}{71} \) or simplify? Wait, no, 1562=2×11×71, so no square factors. Then \( \tan(G)=\frac{\text{opposite}}{\text{adjacent}}=\frac{EF}{FG}=\frac{7}{\sqrt{22}} \). Rationalize: \( \frac{7\sqrt{22}}{22} \).

Wait, let's recheck:

• \( \sin(G) \): opposite side to \( G \) is \( EF = 7 \), hypotenuse \( EG=\sqrt{71} \), so \( \sin(G)=\frac{7}{\sqrt{71}}=\frac{7\sqrt{71}}{71} \).

• \( \cos(G) \): adjacent side to \( G \) is \( FG=\sqrt{22} \), hypotenuse \( EG=\sqrt{71} \), so \( \cos(G)=\frac{\sqrt{22}}{\sqrt{71}}=\frac{\sqrt{22}\times\sqrt{71}}{71}=\frac{\sqrt{1562}}{71} \)? Wait, no, \( \sqrt{22}\times\sqrt{71}=\sqrt{22\times71}=\sqrt{1562} \), but 1562=2×11×71, so can't simplify further. Alternatively, maybe I made a mistake in hypotenuse? Wait, \( EF = 7 \), \( FG=\sqrt{22} \), right angle at F, so yes, \( EG=\sqrt{7^2 + (\sqrt{22})^2}=\sqrt{49 + 22}=\sqrt{71} \), correct.

• \( \tan(G) \): opposite over adjacent, \( \frac{EF}{FG}=\frac{7}{\sqrt{22}}=\frac{7\sqrt{22}}{22} \) (rationalizing by multiplying numerator and denominator by \( \sqrt{22} \)).

Answer:

\( \sin(G) = \boldsymbol{\frac{7\sqrt{71}}{71}} \)

\( \cos(G) = \boldsymbol{\frac{\sqrt{1562}}{71}} \) (or \( \frac{\sqrt{22}\sqrt{71}}{71} \))

\( \tan(G) = \boldsymbol{\frac{7\sqrt{22}}{22}} \)

Wait, actually, \( \sqrt{22}\times\sqrt{71}=\sqrt{22\times71}=\sqrt{1562} \), but 1562=2×11×71, so no square factors. Alternatively, maybe the problem expects leaving it as \( \frac{\sqrt{22}}{\sqrt{71}} \) but rationalized, so \( \frac{\sqrt{1562}}{71} \) is correct, or \( \frac{\sqrt{22}\sqrt{71}}{71} \). But maybe I made a mistake in hypotenuse? Wait, 7 squared is 49, sqrt(22) squared is 22, 49+22=71, so hypotenuse is sqrt(71), correct. So:

• \( \sin(G) = \frac{7}{\sqrt{71}} = \frac{7\sqrt{71}}{71} \) (rationalized)

• \( \cos(G) = \frac{\sqrt{22}}{\sqrt{71}} = \frac{\sqrt{22}\times\sqrt{71}}{71} = \frac{\sqrt{1562}}{71} \) (rationalized)

• \( \tan(G) = \frac{7}{\sqrt{22}} = \frac{7\sqrt{22}}{22} \) (rationalized)