Exercise 1

Step 1

1 of 3

a) The highest number must be the hypotenuse from Pythagorean equation $c^2=a^2+b^2$.

The greater the measure of an angle, the longer is its opposite side.

Since one side is $30^circ$, the other angle must be $90-30=60^circ$

So the side opposite to $30^circ$ must be the smallest side which is 1.

Exercise scan

Step 2

2 of 3

b) With respect to $30^circ$, the adjacent side is $2$ and the opposite side is $1$

c) With respect to $60^circ$, the adjacent side is $1$ and the opposite side is $2$

Result

3 of 3

a) see sketch inside

b) adjacent side: 2 ; opposite side: 1

c) adjacent side: 1 ; opposite side: 2

Exercise 2

Step 1

1 of 3

a) The longest side must be the hypotenuse, and the others are the legs. Exercise scan

Step 2

2 of 3

b) The opposite side with respect to any of the $45^circ$ is 1 and the adjacent side is also 1.

A triangle with equal lengths of legs is called an isosceles right triangle.

Result

3 of 3

a) see sketch inside

b) both opposite and adjacent side are equal to 1

Exercise 3

Step 1

1 of 3

Evaluating Trigo function

begin{table}[]
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 $alpha$end{tabular}} \ hline
multicolumn{1}{|l|}{$0leq theta<90^circ$} & 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 angleend{tabular}} & multicolumn{1}{l|}{$cos alpha$} & multicolumn{1}{l|}{$sin alpha$} & \ 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}

Step 2

2 of 3

All are acute angles in quadrant I, hence, refer to the table of special angles. Otherwise, you have to find the reference acute angle based on the first table. We suggest you memorize the values in the second table as this will come up often in the future.

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

b) $cos 30^circ = dfrac{sqrt{3}}{2}$

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

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

Result

3 of 3

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

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

c) $1$

d) $dfrac{sqrt{2}}{2}$

Exercise 4

Step 1

1 of 3

Evaluating Trigo function

begin{table}[]
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 $alpha$end{tabular}} \ hline
multicolumn{1}{|l|}{$0leq theta<90^circ$} & 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 angleend{tabular}} & multicolumn{1}{l|}{$cos alpha$} & multicolumn{1}{l|}{$sin alpha$} & \ 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}

Exercise scan

Step 2

2 of 3

All are acute angles in quadrant I, hence, refer to the table of special angles. Otherwise, you have to find the reference acute angle based on the first table. We suggest you memorize the values in the second table as this will come up often in the future.

Remember that $tan theta = dfrac{sin theta}{cos theta}$

a) $sin 30^circ times tan 60^circ -cos 30^circ=dfrac{1}{2}times dfrac{sqrt{3}/2}{1/2}-dfrac{sqrt{3}}{2}=0$

b) $2cos 45^circ times sin 45^circ=2times dfrac{sqrt{2}}{2}times dfrac{sqrt{2}}{2}=1$

c) $tan^230^circ-cos^245^circ=left(dfrac{1/2}{sqrt{3}/2}right)^2-left(dfrac{sqrt{3}}{2}right)^2=-dfrac{1}{6}$

d) $1-dfrac{sin 45^circ}{cos 45^circ}=1-dfrac{sqrt{2}/2}{sqrt{2}/2}=0$

Result

3 of 3

a) 0
b) 1
c) $-1/6$
d) 0

Exercise 5

Step 1

1 of 3

Evaluating Trigo function

begin{table}[]
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 $alpha$end{tabular}} \ hline
multicolumn{1}{|l|}{$0leq theta<90^circ$} & 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 angleend{tabular}} & multicolumn{1}{l|}{$cos alpha$} & multicolumn{1}{l|}{$sin alpha$} & \ 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}

Exercise scan

Step 2

2 of 3

All are acute angles in quadrant I, hence, refer to the table of special angles. Otherwise, you have to find the reference acute angle based on the first table. We suggest you memorize the values in the second table as this will come up often in the future.\\
[begin{gathered}
{text{a}}){ ;sin ^2}{30^{text{o}}} + {cos ^2}{30^{text{o}}} = {left( {frac{1}{2}} right)^2} + {left( {frac{{sqrt 3 }}{2}} right)^2} = frac{1}{4} + frac{3}{4} = 1 hfill \
hfill \
{text{b) }}{sin ^2}{45^{text{o}}} + {cos ^2}{45^{text{o}}} = {left( {frac{{sqrt 2 }}{2}} right)^2} + {left( {frac{{sqrt 2 }}{2}} right)^2} = frac{2}{4} + frac{2}{4} = 1 hfill \
hfill \
{text{c) si}}{{text{n}}^2}{60^{text{o}}} + {cos ^2}{60^{text{o}}} = {left( {frac{{sqrt 3 }}{2}} right)^2} + {left( {frac{1}{2}} right)^2} = frac{3}{4} + frac{1}{4} = 1 hfill \
end{gathered} ]

Result

3 of 3

See answer inside.

Exercise 6

Step 1

1 of 5

Step 2

2 of 5

Refer to the angle $30^circ$ on the special triangles.

a) $sin 30^circ = dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{1/2}{1}=dfrac{1}{2}$

$cos 30^circ =dfrac{text{adjacent side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}=dfrac{sqrt{3}}{2}$

$tan 30^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}$

$dfrac{sin 30^circ}{cos 30^circ}=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}$

$therefore dfrac{sin theta}{cos theta}=tan theta$

Step 3

3 of 5

Refer to the special triangle with $45^circ$

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

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

$tan45^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{1}{1}=1$

$dfrac{sin 45^circ}{cos 45^circ}=dfrac{sqrt{2}/2}{sqrt{3}/2}=1$

$$
therefore dfrac{sin theta}{cos theta}=tan theta
$$

Step 4

4 of 5

Refer to the special triangle with $60^circ$

c) $sin 60^circ = dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}=dfrac{sqrt{3}}{2}$

$cos 60^circ =dfrac{text{adjacent side}}{text{hypotenuse}}=dfrac{1/2}{1}=dfrac{1}{2}$

$tan60^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{sqrt{3}/2}{1/2}=sqrt{3}$

$dfrac{sin 45^circ}{cos 45^circ}=dfrac{sqrt{3}/2}{1/2}=sqrt{3}$

$$
therefore dfrac{sin theta}{cos theta}=tan theta
$$

Result

5 of 5

See answer inside.

Exercise 7

Step 1

1 of 6

Step 2

2 of 6

We need to find the value $theta$ from a special triangle, so we need to work backwards

a) $sin theta = dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}implies theta = 60^circ$

Step 3

3 of 6

b) $sqrt{3}tan theta=1$

$tan theta = dfrac{1}{sqrt{3}}=dfrac{text{opposite side}}{text{adjacent side}}=dfrac{1/2}{sqrt{3}/2}$

This happens when

$$
theta = 30^circ
$$

Step 4

4 of 6

c) $2sqrt{2}cos theta =2$

$cos theta = dfrac{2}{2sqrt{2}}$

$cos theta = dfrac{1}{sqrt{2}}$

$cos theta = dfrac{sqrt{2}}{2}$

This happens when $theta =45^circ$

Step 5

5 of 6

d) $2cos theta =sqrt{3}$

$cos theta = dfrac{sqrt{3}/2}{1}=dfrac{text{adjacent side}}{text{hypotenuse}}$

This happens when $theta=30^circ$

Result

6 of 6

a) $theta = 60^circ$

b) $theta=30^circ$

c) $theta =45^circ$

d) $theta=30^circ$

Exercise 8

Step 1

1 of 3

Sketch the triangle as follows. Exercise scan

Step 2

2 of 3

The height of the top of the ladder from the floor is the adjacent side and we are given with hypotenuse which is the length of the ladder. Thus, we shall use cosine.

$cos theta = dfrac{text{adjacent side}}{text{hypotenuse}}$

$cos 30^circ = dfrac{x}{5}$

$x=5cos 30=5cdot dfrac{sqrt{3}}{2}=dfrac{5sqrt{3}}{2}$ m

Here, we assumed that the floor is flat (not inclined) and the wall is vertically straight up so that the wall is perpendicular to the floor.

Result

3 of 3

$dfrac{5sqrt{3}}{2}$ m , assuming the floor is flat

Exercise 9

Step 1

1 of 3

Step 2

2 of 3

From the special triangle containing $30^circ$,

$tan 30^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}$

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

$sin 30^circ =dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{1/2}{1}$

$cos 30^circ=dfrac{text{adjacent side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}$

Therefore,

$tan 30^circ +dfrac{1}{tan 30^circ}=dfrac{1}{sqrt{3}}+sqrt{3}=dfrac{1+3}{sqrt{3}}=dfrac{4}{sqrt{3}}$

$dfrac{1}{sin 30^circ cdot cos 30^circ}=dfrac{1}{1/2cdot sqrt{3}/2}=dfrac{4}{sqrt{3}}$

$$
implies tan 30^circ+dfrac{1}{tan 30^circ}=dfrac{1}{sin 30^circ cos 30^circ}=dfrac{4}{sqrt{3}}
$$

Result

3 of 3

$$
tan 30^circ+dfrac{1}{tan 30^circ}=dfrac{1}{sin 30^circ cos 30^circ}=dfrac{4}{sqrt{3}}
$$

Exercise 10

Step 1

1 of 3

a) Sarah can form a right triangle such that the hypotenuse is the distance from the home to the second base and the legs are both $27.4$ m. Then, Sarah can use Pythagorean theorem to solve for the hypotenuse.

Exercise scan

Step 2

2 of 3

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

$x^2=sqrt{27.4^2+27.4^2}=38.75$ m

Therefore, the distance from the home to the second base is $38.75$ m

Result

3 of 3

$38.75$ m

Exercise 11

Step 1

1 of 4

Step 2

2 of 4

[begin{gathered}
{text{Refer to the figure in your textbook}}{text{.}} hfill \
hfill \
{text{The area of a triangle is}} hfill \
A = frac{1}{2} cdot left( {{text{base}}} right) cdot left( {{text{altitude}}} right) hfill \
hfill \
{text{The altitude must be perpendicular to the base}}{text{.}} hfill \
hfill \
{text{a) Given that}} hfill \
tan alpha = 1 hfill \
{text{base}} = AC = AB + BC hfill \
{text{height = }}BD = 6 hfill \
hfill \
{text{We shall calculate }}AB{text{ }} hfill \
tan alpha = frac{{{text{opposite side}}}}{{{text{adjacent side}}}} = frac{6}{{AB}} hfill \
AB = frac{6}{{tan alpha }} = frac{6}{1} = 6 hfill \
hfill \
{text{Then calculate }}BC hfill \
hfill \
tan {60^{text{o}}} = frac{{BC}}{{BD}} hfill \
hfill \
BC = 6tan {60^{text{o}}} = 6 cdot frac{{sin {{60}^{text{o}}}}}{{cos {{60}^{text{o}}}}} = left( {frac{{sqrt 3 /2}}{{1/2}}} right) hfill \
BC = 6sqrt 3 hfill \
hfill \
{text{Therefore, }} hfill \
AC = AB + BD = 6 + 6sqrt 3 hfill \
hfill \
{text{Area}} = frac{1}{2}left( {AC} right)left( {BD} right) = frac{1}{2}left( {6 + sqrt 3 } right)left( 6 right) hfill \
{text{Area}} = 3left( {6 + sqrt 3 } right) = 18 + 3sqrt 3 hfill \
hfill \
hfill \
end{gathered} ]

Step 3

3 of 4

[begin{gathered}
{text{b) Given that}} hfill \
cos beta = frac{{sqrt 3 }}{2} Rightarrow beta = {30^{text{o}}} hfill \
{text{base}} = SQ = SR + RQ hfill \
{text{height = }}PR hfill \
hfill \
{text{We shall calculate }}SR hfill \
sin beta = frac{{{text{opposite side}}}}{{{text{hypotenuse}}}} = frac{{SR}}{{PS}} hfill \
sin {30^{text{o}}} = frac{{SR}}{{13}} Rightarrow SR = 13sin {30^{text{o}}} hfill \
SR = 13left( {frac{1}{2}} right) = frac{{13}}{2} hfill \
hfill \
{text{Then we shall calculate PR}} hfill \
{text{cos}}beta = frac{{{text{adjacent side}}}}{{{text{hypotenuse}}}} hfill \
cos beta = frac{{PR}}{{13}} hfill \
PR = 13left( {frac{{sqrt 3 }}{2}} right) = frac{{13sqrt 3 }}{2} hfill \
{text{Notice that since }}angle Q = {45^{text{o}}}{text{ so the}} hfill \
{text{its corresponding legs are equal, thus}} hfill \
QR = PR = frac{{13sqrt 3 }}{2} hfill \
{text{So now we have}} hfill \
{text{base}} = SR + QR = frac{{13}}{2} + frac{{13sqrt 3 }}{2} = frac{{13}}{2}left( {1 + sqrt 3 } right) hfill \
{text{height = }}PR = frac{{13sqrt 3 }}{2} hfill \
hfill \
{text{Area = }}frac{1}{2}left( {{text{base}}} right)left( {{text{height}}} right) hfill \
{text{Area}} = frac{1}{2}left[ {frac{{13}}{2}left( {1 + sqrt 3 } right)} right] times frac{{13sqrt 3 }}{2} hfill \
{text{Area}} = frac{{168sqrt 3 }}{8}left( {1 + sqrt 3 } right) hfill \
end{gathered} ]

Result

4 of 4

a) $18+3sqrt{3}$

b) $dfrac{168sqrt{3}}{8}(1+sqrt{3})$

Exercise 12

Step 1

1 of 5

Use your calculator to evaluate the expression

a) $2.595$

Exercise scan

Step 2

2 of 5

b) Here, we need to express in radical form. Thus, we shall use the special right triangles

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

$cos 30^circ = dfrac{text{adjacent side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}=dfrac{sqrt{3}}{2}$

$tan 60^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{sqrt{3}/2}{1/2}=sqrt{3}$

$sin 60^circ =dfrac{text{opposite side}}{text{hypotenuse}}=dfrac{sqrt{3}/2}{1}=dfrac{sqrt{3}}{2}$

$$
tan 30^circ = dfrac{text{opposite side}}{text{adjacent side}}=dfrac{1/2}{sqrt{3}/2}=dfrac{1}{sqrt{3}}
$$

Step 3

3 of 5

[begin{gathered}
sin {45^{text{o}}}left( {1 – cos {{30}^{text{o}}}} right) + 5tan {60^{text{o}}}left( {sin {{60}^{text{o}}} – tan {{30}^{text{o}}}} right) hfill \
= frac{{sqrt 2 }}{2}left( {1 – frac{{sqrt 3 }}{2}} right) + 5left( {sqrt 3 } right)left( {frac{{sqrt 3 }}{2} – frac{{sqrt 3 }}{3}} right) hfill \
= frac{{sqrt 2 }}{2} – frac{{sqrt 6 }}{4} + 5left( {frac{3}{2} – frac{3}{3}} right) hfill \
= frac{{sqrt 2 }}{2} – frac{{sqrt 6 }}{4} + 5left( {frac{1}{2}} right) hfill \
= frac{{sqrt 2 }}{2} – frac{{sqrt 6 }}{4} + frac{5}{2} hfill \
end{gathered} ]

Step 4

4 of 5

c) The answer obtained from the calculator is not the exact value but just an approximation.

Result

5 of 5

a) 2.595

b) $dfrac{sqrt{2}}{2}-dfrac{sqrt{6}}{4}+dfrac{5}{2}$

c) calculator answer is just an approximation

Exercise 13

Step 1

1 of 3

Step 2

2 of 3

[begin{gathered}
{text{cot}},alpha = sqrt 3 hfill \
Rightarrow tan alpha = frac{1}{{sqrt 3 }} = alpha = {tan ^{ – 1}}left( {frac{1}{{sqrt 3 }}} right) hfill \
alpha = {30^{text{o}}} hfill \
{text{From the special triangles,}} hfill \
{text{}} hfill \
sin {30^{text{o}}} = frac{1}{2} hfill \
cos 3{0^{text{o}}} = frac{{sqrt 3 }}{2} hfill \
hfill \
left( {sin alpha } right)left( {cot alpha } right) – {cos ^2}alpha hfill \
= left( {frac{1}{2}} right)left( {sqrt 3 } right) – {left( {frac{{sqrt 3 }}{2}} right)^2} hfill \
= frac{{sqrt 3 }}{2} – frac{3}{4} hfill \
hfill \
end{gathered} ]

Result

3 of 3

$$
dfrac{sqrt{3}}{2}-dfrac{3}{4}
$$

All Solutions

Section 5-2: Evaluating Trigonometric Ratios for Special Angles

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