Question

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

right triangle qsr with right angle at s, qs = $3\sqrt{5}$, sr = $\sqrt{33}$

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

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

Explanation:

Step1: Find the hypotenuse \( QR \)

In right triangle \( QSR \), by Pythagorean theorem \( QR^2 = QS^2 + SR^2 \). Given \( QS = 3\sqrt{5} \), \( SR=\sqrt{33} \).
\[ \begin{align*} QR^2&=(3\sqrt{5})^2+(\sqrt{33})^2\\ &= 9\times5 + 33\\ &= 45+33\\ &= 78 \end{align*} \]
So \( QR=\sqrt{78} = \sqrt{6\times13}= \sqrt{6}\times\sqrt{13}\) (but we can keep it as \(\sqrt{78}\) for now).

Step2: Calculate \( \sin(Q) \)

\(\sin(Q)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{SR}{QR}\)
\[ \begin{align*} \sin(Q)&=\frac{\sqrt{33}}{\sqrt{78}}\\ &=\frac{\sqrt{33}}{\sqrt{6\times13}}\\ &=\frac{\sqrt{33}}{\sqrt{6}\times\sqrt{13}}\\ &=\frac{\sqrt{33}\times\sqrt{6}}{6\times\sqrt{13}}\\ &=\frac{\sqrt{198}}{6\sqrt{13}}\\ &=\frac{\sqrt{9\times22}}{6\sqrt{13}}\\ &=\frac{3\sqrt{22}}{6\sqrt{13}}\\ &=\frac{\sqrt{22}}{2\sqrt{13}}\\ &=\frac{\sqrt{22}\times\sqrt{13}}{2\times13}\\ &=\frac{\sqrt{286}}{26} \end{align*} \]
Wait, maybe a simpler way: simplify \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36...}} \) no, better to simplify \( \sqrt{78}=\sqrt{3\times26}=\sqrt{3\times2\times13} \), \( \sqrt{33}=\sqrt{3\times11} \). So \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{3\times11}}{\sqrt{3\times26}}=\frac{\sqrt{11}}{\sqrt{26}}=\frac{\sqrt{11}\times\sqrt{26}}{26}=\frac{\sqrt{286}}{26} \) (same as before). Wait, maybe I made a mistake in Pythagorean theorem. Wait \( (3\sqrt{5})^2 = 9*5=45 \), \( (\sqrt{33})^2=33 \), 45 +33=78, correct.

Wait, alternatively, let's re - express:

\(\sin(Q)=\frac{SR}{QR}=\frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36}} \) no, wait 78 = 33 + 45, yes. Wait, maybe factor 78 and 33: 33 = 3*11, 78=3*26. So \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{3\times11}}{\sqrt{3\times26}}=\frac{\sqrt{11}}{\sqrt{26}}=\frac{\sqrt{11}\times\sqrt{26}}{26}=\frac{\sqrt{286}}{26} \)

Step3: Calculate \( \cos(Q) \)

\(\cos(Q)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{QS}{QR}\)
\[ \begin{align*} \cos(Q)&=\frac{3\sqrt{5}}{\sqrt{78}}\\ &=\frac{3\sqrt{5}}{\sqrt{6\times13}}\\ &=\frac{3\sqrt{5}\times\sqrt{6}}{6\sqrt{13}}\\ &=\frac{3\sqrt{30}}{6\sqrt{13}}\\ &=\frac{\sqrt{30}}{2\sqrt{13}}\\ &=\frac{\sqrt{30}\times\sqrt{13}}{2\times13}\\ &=\frac{\sqrt{390}}{26} \end{align*} \]
Alternatively, \( \frac{3\sqrt{5}}{\sqrt{78}}=\frac{3\sqrt{5}}{\sqrt{5\times15.6}} \) no, 78 = 5*15.6 no, 78=5*15.6 is wrong. 78 divided by 5 is 15.6, no. Wait 78 = 3*26, 5*15.6 is not helpful. Wait, 3\sqrt{5} over \sqrt{78} can be rationalized as \( \frac{3\sqrt{5}\times\sqrt{78}}{78}=\frac{3\sqrt{390}}{78}=\frac{\sqrt{390}}{26} \) (since 390 = 5*78, 78=3*26, so \(\sqrt{390}=\sqrt{5\times3\times26}\))

Step4: Calculate \( \tan(Q) \)

\(\tan(Q)=\frac{\text{opposite}}{\text{adjacent}}=\frac{SR}{QS}\)
\[ \begin{align*} \tan(Q)&=\frac{\sqrt{33}}{3\sqrt{5}}\\ &=\frac{\sqrt{33}\times\sqrt{5}}{3\times5}\\ &=\frac{\sqrt{165}}{15} \end{align*} \]

Wait, maybe I messed up the hypotenuse calculation. Wait, let's check again: \( QS = 3\sqrt{5} \), \( SR=\sqrt{33} \), right triangle at S. So \( QR^2=QS^2 + SR^2=(3\sqrt{5})^2+(\sqrt{33})^2 = 45 + 33=78 \), correct. So \( QR=\sqrt{78} \)

Now, \( \sin(Q)=\frac{SR}{QR}=\frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36...}} \) no, \( \sqrt{78}=\sqrt{33 + 45}=\sqrt{33+ (3\sqrt{5})^2} \), correct.

Wait, another approach: simplify \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36}} \) no, 78 = 33*2.36 is not an integer. Wait, 78 = 33*2 + 12? No, 33*2=66, 78-66=12. No, 78= 2*39=2*3*13, 33=3*11. So \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{3\times11}}{\sqrt{2\times3\times13}}=\frac{\sqrt{11}}{\sqrt{2\times13}}=\frac{\sqrt{11}}{\sqrt{26}}=\frac{\sqrt{11}\times\sqrt{26}}{26}=\frac{\sqrt{286}}{26} \)

\( \cos(Q)=\frac{QS}{QR}=\frac{3\sqrt{5}}{\sqrt{78}}=\frac{3\sqrt{5}}{\sqrt{2\times3\times13}}=\frac{3\sqrt{5}\times\sqrt{2\times3\times13}}{78}=\frac{3\sqrt{390}}{78}=\frac{\sqrt{390}}{26} \) (since 390=5*78, and 78=2*3*13, so \(\sqrt{390}=\sqrt{5\times2\times3\times13}\))

\( \tan(Q)=\frac{SR}{QS}=\frac{\sqrt{33}}{3\sqrt{5}}=\frac{\sqrt{33}\times\sqrt{5}}{3\times5}=\frac{\sqrt{165}}{15} \) (because \( \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b} \), so \( \frac{\sqrt{33}}{3\sqrt{5}}=\frac{\sqrt{33}\times\sqrt{5}}{3\times5}=\frac{\sqrt{165}}{15} \))

Wait, let's verify \( \tan(Q) \): opposite side to \( Q \) is \( SR=\sqrt{33} \), adjacent side is \( QS = 3\sqrt{5} \), so \( \tan(Q)=\frac{\sqrt{33}}{3\sqrt{5}} \), rationalizing the denominator: multiply numerator and denominator by \( \sqrt{5} \), we get \( \frac{\sqrt{33}\times\sqrt{5}}{3\times5}=\frac{\sqrt{165}}{15} \), correct.

For \( \sin(Q) \): \( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36}} \) no, wait 78 = 33 + 45, and 45=(3\sqrt{5})^2, so \( \sqrt{78}=\sqrt{(3\sqrt{5})^2+(\sqrt{33})^2} \), and \( \sin(Q)=\frac{\sqrt{33}}{\sqrt{78}} \), we can also simplify \( \sqrt{78}=\sqrt{6\times13} \), \( \sqrt{33}=\sqrt{3\times11} \), so there is no common factor, so we rationalize:

\( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}\times\sqrt{78}}{78}=\frac{\sqrt{33\times78}}{78}=\frac{\sqrt{2574}}{78} \), but 2574=33*78=33*33*2.36... no, 33*78=33*(33 + 45)=33^2+33*45=1089+1485=2574, and 2574=9*286, so \( \sqrt{2574}=3\sqrt{286} \), so \( \frac{3\sqrt{286}}{78}=\frac{\sqrt{286}}{26} \), correct.

For \( \cos(Q) \): \( \frac{3\sqrt{5}}{\sqrt{78}}=\frac{3\sqrt{5}\times\sqrt{78}}{78}=\frac{3\sqrt{390}}{78}=\frac{\sqrt{390}}{26} \), and 390=10*39=10*3*13, so no further simplification.

Wait, maybe I made a mistake in the Pythagorean theorem. Wait, let's check the lengths again. If \( QS = 3\sqrt{5}\approx3*2.236 = 6.708 \), \( SR=\sqrt{33}\approx5.744 \), then hypotenuse \( QR\approx\sqrt{6.708^2 + 5.744^2}\approx\sqrt{45 + 33}=\sqrt{78}\approx8.832 \), correct.

Now, \( \sin(Q)=\frac{SR}{QR}\approx\frac{5.744}{8.832}\approx0.65 \), and \( \frac{\sqrt{286}}{26}\approx\frac{16.91}{26}\approx0.65 \), correct.

\( \cos(Q)=\frac{QS}{QR}\approx\frac{6.708}{8.832}\approx0.76 \), and \( \frac{\sqrt{390}}{26}\approx\frac{19.75}{26}\approx0.76 \), correct.

\( \tan(Q)=\frac{SR}{QS}\approx\frac{5.744}{6.708}\approx0.856 \), and \( \frac{\sqrt{165}}{15}\approx\frac{12.845}{15}\approx0.856 \), correct.

So:

\( \sin(Q)=\frac{\sqrt{286}}{26} \)

\( \cos(Q)=\frac{\sqrt{390}}{26} \)

\( \tan(Q)=\frac{\sqrt{165}}{15} \)

Wait, but maybe there is a simpler way. Let's check the values again. Wait, 78 = 6*13, 33=3*11, 45=9*5.

Wait, another approach: Let's find the hypotenuse correctly. \( QS = 3\sqrt{5} \), \( SR=\sqrt{33} \). Then \( QR^2=(3\sqrt{5})^2+(\sqrt{33})^2 = 45 + 33 = 78 \), so \( QR=\sqrt{78} \).

Now, \( \sin Q=\frac{SR}{QR}=\frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}}{\sqrt{33\times2.36}} \) no, wait 78 = 33*2.36 is not an integer. Wait, 78 = 33 + 45, and 45 = (3\sqrt{5})^2, so \( \sqrt{78}=\sqrt{(3\sqrt{5})^2+(\sqrt{33})^2} \), and \( \sin Q=\frac{\sqrt{33}}{\sqrt{78}} \), we can also write \( \sqrt{78}=\sqrt{6\times13} \), \( \sqrt{33}=\sqrt{3\times11} \), so there is no common factor, so we rationalize:

\( \frac{\sqrt{33}}{\sqrt{78}}=\frac{\sqrt{33}\times\sqrt{78}}{78}=\frac{\sqrt{33\times78}}{78}=\frac{\sqrt{2574}}{78} \), and 2574 = 9×286, so \( \sqrt{2574}=3\sqrt{286} \), so \( \frac{3\sqrt{286}}{78}=\frac{\sqrt{286}}{26} \), correct.

For \( \cos Q=\frac{QS}{QR}=\frac{3\sqrt{5}}{\sqrt{78}}=\frac{3\sqrt{5}\times\sqrt{78}}{78}=\frac{3\sqrt{390}}{78}=\frac{\sqrt{390}}{26} \), and 390 = 10×39 = 10×3×13, so no further simplification.

For \( \tan Q=\frac{SR}{QS}=\frac{\sqrt{33}}{3\sqrt{5}}=\frac{\sqrt{33}\times\sqrt{5}}{3\times5}=\frac{\sqrt{165}}{15} \), correct.

Answer:

\( \sin(Q)=\boldsymbol{\frac{\sqrt{286}}{26}} \)

\( \cos(Q)=\boldsymbol{\frac{\sqrt{390}}{26}} \)

\( \tan(Q)=\boldsymbol{\frac{\sqrt{165}}{15}} \)