Search

Text Color

Margin Size

Font Type

Enable Dyslexic Font

11.1: The Trigonometric Ratios

Last updated

Mar 4, 2023
Save as PDF
- 11.0: Triangles and Circles
- 11.2: Laws of Sines and Cosines

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

2.1 Side and Angle Relationships

Homework 2.1

1. The sum of the angles is not $180^{\circ}$

3. The exterior angle is not equal to the sum of the opposite interior angles.

5. The sum of the acute angles is not $90^{\circ}$

7. The largest side is not opposite the largest angle.

9. The Pythagorean theorem is not satisfied.

11. $5^2+12^2=13^2$ , but the angle opposite the side of length 13 is $85^{\circ}$ .

13. $4 < x < 16$

15. $0 < x < 16$

17. 21 in

19. $6\sqrt{2}$ in

21. $w = 6\sqrt{10}$ in

23. 29

25. $\sqrt{3}$

27. No

29. Yes

31. No

33. The distance from $(0,0)$ to $(3,3)$ is $3 \sqrt{2}$ , and the distance from $(3,3)$ to $(6,0)$ is also $3 \sqrt{2}$ , so the triangle is isosceles. The distance from $(0,0)$ to $(6,0)$ is 6 , and $(3 \sqrt{2})^2+(3 \sqrt{2})^2=6^2$ so the triangle is a right triangle.

35. 25 ft

37. $\alpha=30^{\circ}, \beta=60^{\circ}, h=\sqrt{3}$

39. $8 \sqrt{3}$ in

41.

a No

b Yes

43.

a $(-1,0)$ and $(1,0) ; 2$

b $\sqrt{(p+1)^2+q^2}$ and $\sqrt{(p-1)^2+q^2}$

$\begin{aligned} \left(\sqrt{(p+1)^2+q^2}\right)^2 & +\left(\sqrt{(p-1)^2+q^2}\right)^2 \\ & =p^2+2 p+1+q^2+p^2-2 p+1+q^2 \\ & =2 p^2+2+2 q^2=2+2\left(p^2+q^2\right) \\ & =2+2(1)=4 \end{aligned}$

2.2 Right Triangle Trigonometry

Homework 2.2

a 0.91

b 0.91

c 0.9063

a 0.77

b 0.77

c 0.7660

a $4\sqrt{13} \approx 14.42$

b \(\sin \theta = 0.5547, \cos \theta = 0.8321, \tan \theta = 0.6667

a $4\sqrt{15} \approx 15.49$

b $\sin \theta = 0.9682, \cos \theta = 0.2500, \tan \theta = 3.8730$

a $2\sqrt{67} \approx 15.49$

b $\sin \theta = 0.2116, \cos \theta = 0.9774, \tan \theta = 0.2165$

11.

$Screen Shot 2023-02-08 at 11.44.49 PM.png$

(Answers may vary)

13.

$Screen Shot 2023-02-08 at 11.44.59 PM.png$

(Answers may vary)

15.

$Screen Shot 2023-02-08 at 11.45.09 PM.png$

(Answers may vary)

17. 14.41

19. 37.86

21. 86.08

23.

$Screen Shot 2023-02-08 at 11.47.46 PM.png$

25.

$Screen Shot 2023-02-08 at 11.47.55 PM.png$

27.

a $Screen Shot 2023-02-08 at 11.48.06 PM.png$

b $\tan 54.8^{\circ} = \dfrac{h}{20}, 170.1$ yd

29.

a $Screen Shot 2023-02-08 at 11.52.25 PM.png$

b $\tan 36.2^{\circ} = \dfrac{260}{d}, 355.2$ ft

31.

a $Screen Shot 2023-02-08 at 11.52.37 PM.png$

b $\sin 48^{\circ} = \dfrac{a}{1500}, 1114.7$ m

33.

a $Screen Shot 2023-02-08 at 11.52.46 PM.png$

b $\cos 38^{\circ} = \dfrac{1800}{x}, 2284.2$ m

35. $x = \dfrac{82}{\tan \theta}$

37. $x = 11 \sin \theta$

39. $x = \dfrac{9}{\cos \theta}$

41. $36 \sin 25^{\circ} \approx 15.21$

43. $46 \sin 20^{\circ} \approx 15.73$

45. $12 \sin 40^{\circ} \approx 7.71$

47.

	$\sin$	$\cos$	$\tan$
$\theta$	$\dfrac{3}{5}$	$\dfrac{4}{5}$	$\dfrac{3}{4}$
$\phi$	$\dfrac{4}{5}$	$\dfrac{3}{5}$	$\dfrac{4}{3}$

49.

	$\sin$	$\cos$	$\tan$
$\theta$	$\dfrac{1}{\sqrt{5}}$	$\dfrac{2}{\sqrt{5}}$	$\dfrac{1}{2}$
$\phi$	$\dfrac{2}{\sqrt{5}}$	$\dfrac{1}{\sqrt{5}}$	2

51.

a $\theta$ and $\phi$ are complements.

b $\sin \theta=\cos \phi$ and $\cos \theta=\sin \phi$ . The side opposite $\theta$ is the side adjacent to $\phi$ , and vice versa.

53.

a As $\theta$ increases, $\tan \theta$ increases also. The side opposite $\theta$ increases in length while the side adjacent to $\theta$ remains fixed.

b As $\theta$ increases, $\cos \theta$ decreases. The side adjacent to $\theta$ remains fixed while the hypotenuse increases in length.

55. As $\theta$ decreases toward $0^{\circ}$ , the side opposite $\theta$ approaches a length of 0, so $\sin \theta$ approaches 0. But as $\theta$ increases toward $90^{\circ}$ , the length of the side opposite $\theta$ approaches the length of the hypotenuse, so $\sin \theta$ approaches 1.

57. The triangle is not a right triangle.

59. $\dfrac{21}{20}$ is the ratio of hypotenuse to the adjacent side, which is the reciprocal of $\cos \theta$ .

61.

a $0.2358$

b sine

c $48^{\circ}$

d $77^{\circ}$

63.

a $\dfrac{5}{12}$

b 3

c $\dfrac{2}{3}$

d $\dfrac{2}{\sqrt{7}}$

65. Although the triangles may differ in size, the ratio of the side adjacent to the angle to the hypotenuse of the triangle remains the same because the triangles would all be similar, and hence corresponding sides are proportional.

67.

a $\dfrac{2}{3}$

b $\dfrac{2}{3}$

c $Screen Shot 2023-02-09 at 12.08.10 AM.png$

2.3 Solving Right Triangles

Homework 2.3

1. $A = 61^{\circ}, a = 25.26, c = 28.88$

3. $A = 68^{\circ}, a = 0.93, b = 0.37$

a $Screen Shot 2023-02-09 at 12.14.20 AM.png$

b $B = 48^{\circ}, a = 17.4, b = 19.3$

a $Screen Shot 2023-02-09 at 12.14.27 AM.png$

b $A=57^{\circ}, b=194.4, c=357.7$

a $Screen Shot 2023-02-09 at 12.14.38 AM.png$

b $B=78^{\circ}, b=18.8, c=19.2$

11.

a $Screen Shot 2023-02-09 at 12.14.48 AM.png$

Solve $\sin 53.7^{\circ} = \dfrac{8.2}{c}$ for $c$ .
Solve $\tan 53.7^{\circ} = \dfrac{8.2}{a}$ for $a$ .
Subtract $53.7^{\circ}$ from $90^{\circ}$ to find $A$ .

13.

a $Screen Shot 2023-02-09 at 12.15.00 AM.png$

Solve $\cos 25^{\circ} = \dfrac{40}{c}$ for $c$ .
Solve $\tan 25^{\circ} = \dfrac{a}{40}$ for $a$ .
Subtract $25^{\circ}$ from $90^{\circ}$ to find $B$ .

15.

a $Screen Shot 2023-02-09 at 12.19.06 AM.png$

Solve $\sin 64.5^{\circ} = \dfrac{a}{24}$ for $a$ .
Solve $\cos 64.5^{\circ} = \dfrac{b}{24}$ for $b$ .
Subtract $64.5^{\circ}$ from $90^{\circ}$ to find $B$ .

17. $74.2^{\circ}$

19. $56.4^{\circ}$

21. $66.0^{\circ}$

23. $11.5^{\circ}$

$Screen Shot 2023-02-09 at 12.19.16 AM.png$

25. $56.3^{\circ}$

$Screen Shot 2023-02-09 at 12.19.26 AM.png$

27. $73.5^{\circ}$

$Screen Shot 2023-02-09 at 12.19.34 AM.png$

29. $\cos 15^{\circ} = 0.9659$ and $\cos ^{-1} 0.9659 = 15^{\circ}$

31. $\tan 65^{\circ} = 2.1445$ and $\tan ^{-1} 2.1445 = 65^{\circ}$

33. $\sin ^{-1} (0.6) \approx 36.87^{\circ}$ is the angle whose sine is 0.6. $(\sin 6^{\circ})^{-1} \approx 9.5668$ is the reciprocal of $\sin 6^{\circ}$ .

35.

a $Screen Shot 2023-02-09 at 12.23.24 AM.png$

b $\sin \theta = \dfrac{1806}{3(2458)}, 14.6^{\circ}$

37.

a $Screen Shot 2023-02-09 at 12.28.38 AM.png$

b $\tan \theta=\dfrac{32}{10}, 72.6^{\circ}$

39.

a $Screen Shot 2023-02-09 at 12.28.46 AM.png$

b $c=10 \sqrt{10} \approx 31.6, A \approx 34.7^{\circ}, B \approx 55.3^{\circ}$

41.

a $Screen Shot 2023-02-09 at 12.28.57 AM.png$

b $a=\sqrt{256.28} \approx 16.0, A \approx 56.5^{\circ}, B \approx 33.5^{\circ}$

43.

a $Screen Shot 2023-02-09 at 12.29.06 AM.png$

b $\tan ^{-1}\left(\dfrac{26}{30}\right) \approx 40.9^{\circ}, \quad 91 \sqrt{1676} \approx 3612.6 \mathrm{~cm}$

45.

a $Screen Shot 2023-02-09 at 12.29.16 AM.png$

b 6415 km

47.

a $Screen Shot 2023-02-09 at 12.29.28 AM.png$

b 462.9 ft

49. (a) and (b)

51. (a) and (d)

53. $\dfrac{\sqrt{3}}{2} \approx 0.8660$

55. $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.5774$

57. 1.0000

59.

$\theta$	$0^{\circ}$	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$	$90^{\circ}$
$\sin \theta$	0	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	1
$\cos \theta$	1	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$	0
$\tan \theta$	0	$\dfrac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undefined

61.

a smaller

b larger

c larger

63. $a=3 \sqrt{3}, b=3, B=30^{\circ}$

65. $a=b=4 \sqrt{2}, B=45^{\circ}$

67. $e=4, f=4 \sqrt{3}, F=120^{\circ}$

69. $d=2 \sqrt{3}, e=2 \sqrt{2}, f=\sqrt{2}+\sqrt{6}, F=75^{\circ}$

71. \(\a=20, b=20, c=20 \sqrt{2})

73.

a $32 \sqrt{3} \mathrm{~cm}$

b $128 \sqrt{3} \mathrm{sq} \mathrm{cm}$

75.

a $10 \mathrm{sq} \mathrm{cm}$

b $10 \sqrt{2} \mathrm{sq} \mathrm{cm}$

c $10 \sqrt{3} \mathrm{sq} \mathrm{cm}$

2.4 Chapter 2 Summary and Review

Chapter 2 Review Problems

1. If $C>93^{\circ}$ , then $A+B+C>180^{\circ}$

3. If $A<B<58^{\circ}$ , then $A+B+C<180^{\circ}$

5. If $C>50^{\circ}$ , then $A+B+C>180^{\circ}$

7. $Screen Shot 2023-02-09 at 12.47.02 AM.png$

9. $a = 97$

11. $c = 52$

13. Yes

15. $\theta=35.26^{\circ}$

17. No. $a=6, c=10$ or $a=9, c=15$

19.

a $w=86.05$

b $\sin \theta=0.7786, \quad \cos \theta=0.6275, \quad \tan \theta=1.2407$

21.

a $y=16.52$

b $\sin \theta=0.6957, \quad \cos \theta=0.7184, \quad \tan \theta=0.9684$

23. $a = 7.89$

25. $x = 3.57$

27. $b = 156.95$

29. $A=30^{\circ}, a=\dfrac{23 \sqrt{3}}{3}, c=\dfrac{46 \sqrt{3}}{3}$

31. $F=105^{\circ}, d=10 \sqrt{2}, e=20, f=10+10 \sqrt{3}$

33. 3 cm

35. 43.30 cm

37. 15.92 m

39. $114.02 \mathrm{ft}, 37.87^{\circ}$

41.

a $60.26^{\circ}$

b $60.26^{\circ}$

c $m=\dfrac{7}{4}=\tan \theta$

43.

a $c^2$

b $b-a,(b-a)^2$

c $\dfrac{1}{2} a b$

d $4\left(\dfrac{1}{2} a b\right)+(a-b)^2=2 a b+b^2-2 a b+a^2=a^2+b^2$

2.1 Side and Angle Relationships

Homework 2.1

2.2 Right Triangle Trigonometry

Homework 2.2

2.3 Solving Right Triangles

Homework 2.3

2.4 Chapter 2 Summary and Review

Chapter 2 Review Problems

Support Center

How can we help?