$$
csc 20=dfrac{1}{sin20}=2.9238
$$

$$
sec 75=dfrac{1}{cos 75}=3.8637
$$

$$
cot 10=dfrac{1}{tan 10}=5.6713
$$

$$
csc 81=dfrac{1}{sin81}=1.0125
$$

To find $theta$, express the reciprocal ratio in terms of primary trigonometric ratio then use the inverse function. We are restricted for $theta$ within QI so no need to find $theta$ for other quadrants.

a) $cot theta= 0.8701implies tan theta=dfrac{1}{0.8701}$

$theta=tan^{-1}left(dfrac{1}{0.8701}right)=7^circ$

b) $sec theta=4.1011implies cos theta=dfrac{1}{4.1011}$

$theta=cos^{-1}left(dfrac{1}{4.1011}right)=76^circ$

c) $csc theta=1.6406implies sintheta=dfrac{1}{1.6406}$

$theta=sin^{-1}left(dfrac{1}{1.6406}right)=38^circ$

d) $sec theta=2.4312implies cos theta=dfrac{1}{2.4312}$

$theta=cos^{-1}left(dfrac{1}{2.4312}right)=66^circ$

a) $theta=7^circ$

b) $theta=76^circ$

c) $theta=38^circ$

d) $theta=66^circ$

This ratio is more than 1, so it cannot be sine or cosine since both have a range within $-1$ to $1$.

if $tan thetaimplies theta=tan^{-1}left(dfrac{7}{5}right)=54.46^circ$

if $csc theta implies theta=sin^{-1}left(dfrac{5}{7}right)=45.58^circ$

if $sec thetaimplies theta=cos^{-1}left(dfrac{5}{7}right)=44.42^circ$

if $cot thetaimplies theta=tan^{-1}left(dfrac{5}{7}right)=36^circ$

You can sketch the triangle as follows.

Since the opposite side and hypotenuse are involved, we shall use sine.

$sin theta=dfrac{text{opposite side}}{text{hypotenuse}}$

$sin 31^circ=dfrac{8.3}{x}$

$x=dfrac{8.3}{sin 31^circ}=16.1$ m

Since $0.5$ m of the rop is required, the total length of the rope must be

$16.1+0.5=16.6$ m

$16.6$ m

If $csc theta < sec theta$

$dfrac{1}{sin theta}cos theta$

We know that for a unit circle

$x^2+y^2=1implies cos^2theta+sin^2 theta=1$

At $theta=45^circ$, $cos theta=sin theta=dfrac{sqrt{2}}{2}$

So above or below $theta=45^circ$, one ratio must be greater than the other to keep the sum of squares constant.

$cos theta$ is maximum at $theta=0$ while $sin theta$ is maximum at $90^circ$,

So the closer the angle to $90^circ$, the higher is the value of $sin theta$

Therefore, if $sin theta>costheta$ , $theta$ must be within $45^circ < theta <90^circ$

$$
45^circ < theta <90^circ
$$

${bold{Steps; in; Evaluating; Trigonometric; Functions}}$

1) If $theta$ is within the interval $0leq thetaleq 360^circ$, proceed to step 2. If not, find a coterminal angle $alpha$ within the range using the formula

coterminal angle $alpha=theta+360^circ cdot n$ where $n$ is an integer.

2) Determine which quadrant does the terminal point of $theta$ lie. Based on the quadrant, you can find the reference acute angle $beta$ according to the table below.

3) Evaluate the value of the trigonometric function based on the reference acute angle. You should remember its value for special angles such as $30^circ$, $45^circ$ and $60^circ$.

4) Determine the sign of the trigonometric function based on its quadrant.

begin{table}[]
center
defarraystretch{2.4}%
begin{tabular}{llll}
hline
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}principal angle;$theta$end{tabular}} & multicolumn{1}{l|}{Quadrant} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}sign $(cos theta, sin theta)$end{tabular}} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $beta$end{tabular}} \ hline
multicolumn{1}{|l|}{$0< theta<90^circ$} & multicolumn{1}{l|}{I} & multicolumn{1}{l|}{$(+,+)$} & multicolumn{1}{l|}{$theta$} \ hline
multicolumn{1}{|l|}{$90^circ <theta < 180^circ$} & multicolumn{1}{l|}{II} & multicolumn{1}{l|}{$(-,+)$} & multicolumn{1}{l|}{$180^circ-theta$} \ hline
multicolumn{1}{|l|}{$180^circ <theta < 270^circ$} & multicolumn{1}{l|}{III} & multicolumn{1}{l|}{$(-,-)$} & multicolumn{1}{l|}{$theta-180^circ$} \ hline
multicolumn{1}{|l|}{$270^circ < theta <360^circ$} & multicolumn{1}{l|}{IV} & multicolumn{1}{l|}{$(+,-)$} & multicolumn{1}{l|}{$360^circ-theta$} \ hline
& & & \ cline{1-3}
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $beta$ end{tabular}} & multicolumn{1}{l|}{$cos beta$} & multicolumn{1}{l|}{$sin beta$} & \ cline{1-3}
multicolumn{1}{|l|}{30$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{45$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{60$^circ$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & \ cline{1-3}
& & &
end{tabular}
end{table}

All angles here are within $0^circ < theta<90^circ$, so we are always at QI.

a) $sin 60^circ=dfrac{sqrt{3}}{2}$

b) $tan 45^circ=dfrac{sin 45^circ}{cos45^circ}=dfrac{sqrt{2}/2}{sqrt{2}/2}=1$

c) $csc 30^circ =dfrac{1}{sin 30^circ}=dfrac{1}{1/2}=2$

d) $sec 45^circ=dfrac{1}{cos 45^circ}=dfrac{1}{sqrt{2}/2}=dfrac{2}{sqrt{2}}=sqrt{2}$

a)$dfrac{sqrt{3}}{2}$

b) 1

c) 2

d) $sqrt{2}$

a) Refer to the figure in your textbook.

Given that $sin alpha=dfrac{1}{2}$

$sin alpha=dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{BD}{BC}$

$BC=dfrac{14}{sin alpha}=dfrac{14}{0.5}=28$

Using Pythagorean formula $c^2=a^2+b^2$

$CD=sqrt{BC^2-BD^2}=sqrt{28^2-14^2}=sqrt{588}=sqrt{196cdot 3}=14sqrt{3}$

For the smaller triangle

$sin 45^circ=dfrac{14}{AB}$

$AB=dfrac{14}{sin 45^circ}=dfrac{14}{sqrt{2}/2}=14sqrt{2}$

Using Pythagorean formula

$AD=sqrt{AB^2-AD^2}=sqrt{(14sqrt{2})^2-14^2}=14$

$$
AC=AD+CD=14+14sqrt{3}
$$

b) $angle A=45^circ$

$cos 45^circ=dfrac{text{adjacent;side}}{text{hypotenuse}}=dfrac{1}{sqrt{2}}=dfrac{sqrt{2}}{2}$

$sin 45^circ=dfrac{text{opposite;side}}{text{hypotenuse}}=dfrac{1}{sqrt{2}}=dfrac{sqrt{2}}{2}$

$tan 45^circ =dfrac{sin theta}{cos theta}=dfrac{sqrt{2}/2}{sqrt{2}/2}=1$

$sin alpha=dfrac{1}{2}implies alpha=sin^{-1}(0.5)=30^circ$

$angle DBC=90^circ-alpha=90-60=30^circ$

$cos (angle DBC)=dfrac{sqrt{3}}{2}$

$sin (angle DBC)=dfrac{1}{2}$

$$
tan (angle DBC)=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}
$$

a) $AB=14sqrt{2}$

$AC=14+14sqrt{3}$

$AD=14$

$BC=28$

$CD=14sqrt{3}$

b) $cos angle A=dfrac{sqrt{2}}{2}$ ; $sin angle A=dfrac{sqrt{2}}{2}$ ; $tan angle A=1$

$cos(angle DBC)=dfrac{sqrt{3}}{2}$ ; $sin (angle DBC)=dfrac{1}{2}$ ; $tan (angle DBC)=dfrac{sqrt{3}}{3}$

${bold{Steps; in; Evaluating; Trigonometric; Functions}}$

1) If $theta$ is within the interval $0leq thetaleq 360^circ$, proceed to step 2. If not, find a coterminal angle $alpha$ within the range using the formula

coterminal angle $alpha=theta+360^circ cdot n$ where $n$ is an integer.

2) Determine which quadrant does the terminal point of $theta$ lie. Based on the quadrant, you can find the reference acute angle $beta$ according to the table below.

3) Evaluate the value of the trigonometric function based on the reference acute angle. You should remember its value for special angles such as $30^circ$, $45^circ$ and $60^circ$.

4) Determine the sign of the trigonometric function based on its quadrant.

begin{table}[]
center
defarraystretch{2.4}%
begin{tabular}{llll}
hline
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}principal angle;$theta$end{tabular}} & multicolumn{1}{l|}{Quadrant} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}sign $(cos theta, sin theta)$end{tabular}} & multicolumn{1}{l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $beta$end{tabular}} \ hline
multicolumn{1}{|l|}{$0< theta<90^circ$} & multicolumn{1}{l|}{I} & multicolumn{1}{l|}{$(+,+)$} & multicolumn{1}{l|}{$theta$} \ hline
multicolumn{1}{|l|}{$90^circ <theta < 180^circ$} & multicolumn{1}{l|}{II} & multicolumn{1}{l|}{$(-,+)$} & multicolumn{1}{l|}{$180^circ-theta$} \ hline
multicolumn{1}{|l|}{$180^circ <theta < 270^circ$} & multicolumn{1}{l|}{III} & multicolumn{1}{l|}{$(-,-)$} & multicolumn{1}{l|}{$theta-180^circ$} \ hline
multicolumn{1}{|l|}{$270^circ < theta <360^circ$} & multicolumn{1}{l|}{IV} & multicolumn{1}{l|}{$(+,-)$} & multicolumn{1}{l|}{$360^circ-theta$} \ hline
& & & \ cline{1-3}
multicolumn{1}{|l|}{begin{tabular}[c]{@{}l@{}}reference acute angle $beta$ end{tabular}} & multicolumn{1}{l|}{$cos beta$} & multicolumn{1}{l|}{$sin beta$} & \ cline{1-3}
multicolumn{1}{|l|}{30$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{45$^circ$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{2}}{2}$} & \ cline{1-3}
multicolumn{1}{|l|}{60$^circ$} & multicolumn{1}{l|}{$dfrac{1}{2}$} & multicolumn{1}{l|}{$dfrac{sqrt{3}}{2}$} & \ cline{1-3}
& & &
end{tabular}
end{table}

(a) From the sketch, $120^circ$ is in QII
with reference angle $beta=180^circ-120^circ=60^circ$

$sin beta=sin 60^circ=dfrac{sqrt{3}}{2}$

$sin theta>0$ in QI and QII

Therefore, another value of $theta$ would be in QI

$$
theta=60^circ
$$

(b) From the sketch, $225^circ$ is in QIII
with reference angle $beta=225^circ-180^circ=45^circ$

$cos beta=cos 45^circ=dfrac{sqrt{2}}{2}$

$cos theta<0$ in QIII and QII

Therefore, another value of $theta$ would be in QII

In QII: $theta=180^circ-beta=180^circ-45^circ=135^circ$

(c) From the sketch, $330^circ$ is in QIV
with reference angle $beta=360^circ-330^circ=30^circ$

$tan beta=tan 30^circ=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}$

$tantheta<0$ in QIV and QII

Therefore, another value of $theta$ would be in QII

In QII: $theta=180-beta=180-30=150^circ$

(c) From the sketch, $300^circ$ is in QIV
with reference angle $beta=360^circ-300^circ=60^circ$

$cosbeta=cos 60^circ=dfrac{1}{2}$

$tantheta>0$ in QI and QIV

Therefore, another value of $theta$ would be in QI

In QI: $theta=beta=60^circ$

(i) see sketches inside

(ii)
a) $theta=60^circ$

b) $theta=135^circ$

c) $theta=150^circ$

d) $theta=60^circ$

b) Using Pythagorean formula

$r^2=x^2+y^2implies r=sqrt{(-9)^2+4^2}=sqrt{97}$

We can obtain $beta$ using sine

$sin beta = dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{4}{sqrt{97}}$

$beta=sin^{-1}left(dfrac{4}{sqrt{97}}right)=24^circ$

c) Since $theta$ is in QII,

$theta=180^circ-beta=180^circ-24^circ$

$theta=156^circ$

a) see sketch inside

b) $beta=24^circ$

c) $theta=156^circ$

$cos theta=dfrac{4}{5}$ means that $theta$ is either in QI or QIV.

Within $0^circ leq thetaleq 360^circ$, $theta$ can only have to possible values.

Thus, having three angles is impossible and Jeff is wrong.

In fact, the only possible angles are

QI: $theta=cos^{-1}left(dfrac{4}{5}right)=37^circ$

QIV: $theta=360^circ-37^circ=323^circ$

We are restricted for $theta$ within QII

We know that if $(x,y)$ is the terminal point in a unit circle,

then $tan theta=dfrac{y}{x}$ and $cos theta=x$

To find the other trigonometric ratios, use the Pythagorean formula with $r=1$.

$x^2+y^2=1$

Divide both sides by $x^2$

$1+left(dfrac{y}{x}right)^2=left(dfrac{1}{x}right)^2$

$1+(tan theta)^2=left(dfrac{1}{cos theta}right)^2$

$dfrac{1}{cos theta}=pmsqrt{1+left(-dfrac{15}{8}right)^2}=pmsqrt{dfrac{289}{64}}=pmdfrac{17}{8}$

In QII, $cos theta0$, thus,

$sin theta=dfrac{15}{17}$

$sec theta=dfrac{1}{cos theta}=-dfrac{17}{8}$

$csctheta=dfrac{1}{sin theta}=dfrac{17}{15}$

$cot theta=dfrac{1}{tantheta}=-dfrac{8}{15}$

b) $beta=tan^{-1}left(left|dfrac{-15}{8}right|right)=62^circ$

In QII, the principal angle is $theta=180^circ-beta$

$$
theta=180^circ-62^circ=118^circ
$$

a) $cos theta=-dfrac{8}{17}$ ; $sin theta=dfrac{15}{17}$ ; $sectheta=-dfrac{17}{8}$ ; $csctheta=dfrac{17}{15}$ ; $cot theta=-dfrac{8}{15}$

b) $theta=118^circ$

$bold{Steps;in;Solving;Trigonometric;Equations}$\\
1) Find the reference acute angle $beta$ using inverse trigonometric function as\ \
$f(theta)=cimplies beta=f^{-1}(|c|)$\\
2) Based on the sign of $c$, determine the possible quadrants for $theta$ within the given range.\\
3) Find the principal angle $theta$ on those quadrants that has reference acute angle of $beta$
begin{table}[]
defarraystretch{1.4}%
begin{tabular}{|l|l|l|l|l|}
hline
& QI & QII & QIII & QIV\ hline
sine & $+$ & $+$ & $-$ & $-$ \ hline
cosine & $+$ & $-$ & $-$ & $+$ \ hline
tangent & $+$ & $-$ & $+$ & $-$ \ hline
principal angle $theta$ & $;;;;beta;;;;$ & $180^circ-beta$ & $180^circ+beta$ & $360^circ-beta$ \ hline
end{tabular}

end{table}

$sin theta=-0.8190 implies beta=sin^{-1}(|-0.8190|)=55^circ$

$sin theta<0$ in QIII and QIV.

QIII: $theta=180^circ+beta=180^circ+55^circ=235^circ$

QIV: $theta=360^circ-beta=360^circ-55^circ=305^circ$

Therefore

$$
theta=235^circ,;305^circ
$$

$$
theta=235^circ,;305^circ
$$

Without using a calculator, we shall determine which ratios are not possible. We know that

1) $sin theta$ and $cos theta$ must be within $-1leq f(theta)leq1$

2) $csctheta$ and $sectheta$ can NOT take values within $-1<f(theta)<1$.

3) $tantheta$ and $cot theta$ can take any value in $bold{R}$.

Among the given options, it is clear that

a) $cos theta=2.3151$

f) $sin theta=2.3151$

is FALSE because these ratios cannot be more than 1.