11.2: Laws of Sines and Cosines

3.1 Obtuse Angles

Homework 3.1

1.

a \(150^{\circ}\)

b \(135^{\circ}\)

c \(60^{\circ}\)

d \(155^{\circ}\)

e \(25^{\circ}\)

f \(70^{\circ}\)

3.

a (5,2)

b \(\sqrt{29}\)

c \(\cos \theta=\dfrac{5}{\sqrt{29}}, \quad \sin \theta=\dfrac{2}{\sqrt{29}}, \quad \tan \theta=\dfrac{2}{5}\)

5.

a (-4,7)

b \(\sqrt{65}\)

c \(\cos \theta=\dfrac{-4}{\sqrt{65}}, \quad \sin \theta=\dfrac{7}{\sqrt{65}}, \quad \tan \theta=\dfrac{-7}{4}\)

7.

a \(\sin \theta = \dfrac{9}{\sqrt{97}}, \quad \cos \theta = \dfrac{4}{\sqrt{97}}\)

b

c \(\sin \left(180^{\circ}-\theta\right)=\dfrac{9}{\sqrt{97}}, \quad \cos \left(180^{\circ}-\theta\right)=\dfrac{-4}{\sqrt{97}}\)

d \(\theta = 66^{\circ}, \quad 180^{\circ} - \theta = 114^{\circ}\)

9.

a \(\sin \theta=\dfrac{8}{\sqrt{89}}, \quad \cos \theta=\dfrac{-5}{\sqrt{89}}\)

b

c \(\sin \left(180^{\circ}-\theta\right)=\dfrac{8}{\sqrt{89}}, \quad \cos \left(180^{\circ}-\theta\right)= \dfrac{5}{\sqrt{89}}\)

d \(\theta = 122^{\circ}, \quad 180^{\circ} - \theta = 58^{\circ}\)

11.

a

b \(\cos \theta=\dfrac{-5}{13}, \quad \sin \theta=\dfrac{12}{13}, \quad \tan \theta=\dfrac{-12}{5}\)

c \(112.6^{\circ}\)

13.

a

b \(\cos \theta=\dfrac{3}{5}, \quad \tan \theta=\dfrac{-3}{4}\)

c \(143.1^{\circ}\)

15.

a

b \(\sin \theta=\dfrac{\sqrt{112}}{11}, \quad \tan \theta=\dfrac{\sqrt{112}}{3}\)

c \(74.2^{\circ}\)

17.

a

b \(\sin \theta=\dfrac{1}{\sqrt{37}}, \quad \cos \theta=\dfrac{-6}{\sqrt{37}}\)

c \(170.5^{\circ}\)

19.

a

b \(\sin \theta=\dfrac{4}{\sqrt{17}}, \quad \cos \theta=\dfrac{1}{\sqrt{17}}\)

c \(76.0^{\circ}\)

21.

\(\theta\)	\(0^{\circ}\)	\(30^{\circ}\)	\(45^{\circ}\)	\(60^{\circ}\)	\(90^{\circ}\)	\(120^{\circ}\)	\(135^{\circ}\)	\(150^{\circ}\)	\(180^{\circ}\)
\(\cos \theta\)	1	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{\sqrt{2}}\)	\(\dfrac{1}{2}\)	0	\(\dfrac{-1}{2}\)	\(\dfrac{1}{\sqrt{2}}\)	\(\dfrac{-\sqrt{3}}{2}\)	-1
\(\sin \theta\)	0	\(\dfrac{1}{2}\)	\(\dfrac{1}{\sqrt{2}}\)	\(\dfrac{\sqrt{3}}{2}\)	1	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{\sqrt{2}}\)	\(\dfrac{1}{2}\)	0
\(\tan \theta\)	0	\(\dfrac{1}{\sqrt{3}}\)	1	\(\sqrt{3}\)	undefined	\(-\sqrt{3}\)	-1	\(\dfrac{-1}{\sqrt{3}}\)	0

23.

a \(\sin \theta = \sin (180^{\circ} - \theta)\)

b \(\cos \theta = -\cos (180^{\circ} - \theta)\)

c \(\tan \theta = -\tan (180^{\circ} - \theta)\)

25.

a \(\theta \approx 41.4^{\circ}, \quad \phi \approx 138.6^{\circ}\)

b

c \(\sin \theta = \sin \phi = \dfrac{\sqrt{7}}{4}\)

27.

a \(\theta \approx 41.4^{\circ}, \phi \approx 138.6^{\circ}\)

b

c \(\sin \theta = \sin \phi = \dfrac{\sqrt{156279}}{400} \approx 0.9883\)

29. \(44.4^{\circ}\) and \(135.6^{\circ}\)

31. \(57.1^{\circ}\) and \(122.9^{\circ}\)

33. \(41.8^{\circ}\) and \(138.2^{\circ}\)

35. \(\sin 123^{\circ}=q, \quad \cos 33^{\circ}=q, \quad \cos 147^{\circ}=-q\)

37. \(\cos 106^{\circ}=-m, \quad \sin 16^{\circ}=m, \quad \sin 164^{\circ}=m\)

39.

a

b (4,3), (8,6)

с \(y=\tan ^{-1} \dfrac{3}{4} \approx 36.87^{\circ}\)

d

41.

a \(b=8\) in, \(h=3 \sqrt{3}\) in

b \(12 \sqrt{3} \mathrm{sq}\) in

43.

a \(b=6-\dfrac{3 \sqrt{2}}{2} \mathrm{mi}, h=\dfrac{3 \sqrt{2}}{2} \mathrm{mi}\)

b \(\dfrac{18 \sqrt{2}-9}{4}\) sq mi

45.

a \((-1, \sqrt{3})\)

b \((-\sqrt{3}, 3)\)

47.

a \((-3,3)\)

b \((-\sqrt{5}, \sqrt{5})\)

49. 20.71 sq m

51. 55.51 sq cm

53. 6.36 sq in

55. 38.04 sq units

57. 13,851.3 sq ft

59.

a \((-74.97,59.00)\)

b \(B C=141.97, \quad P C=59.00\)

c \(153.74\)

61. \(\dfrac{\sqrt{5}-1}{4}\)

63. Bob found an acute angle. The obtuse angle is the supplement of \(17.46^{\circ}\), or \(162.54^{\circ}\).

65.

a

b \(\cos \theta=\dfrac{x}{3}, \quad \sin \theta=\dfrac{\sqrt{9-x^2}}{3}, \quad \tan \theta=\dfrac{\sqrt{9-x^2}}{x}\)

67.

a

b \(\cos \theta=\dfrac{-\sqrt{4-y^2}}{2}, \sin \theta=\dfrac{y}{2}, \tan \theta=\dfrac{-y}{\sqrt{4-y^2}}\)

69.

a

b \(\cos \theta=\dfrac{-1}{\sqrt{1+m^2}}, \sin \theta= \dfrac{-m}{\sqrt{1+m^2}}, \tan \theta=m\)

3.2 The Law of Sines

Homework 3.2

1. \(x=7.85\)

3. \(q=33.81\)

5. \(d=28.37\)

7. \(\theta=30.80^{\circ}\)

9. \(\theta=126.59^{\circ}\)

11. \(\beta=37.14^{\circ}\)

13. \(a = 4.09, c = 9.48, C = 115^{\circ}\)

15. \(b = 2.98, A = 36.54^{\circ}, B = 99.46^{\circ}\)

17. \(a = 43.55, b = 54.62, C = 99^{\circ}\)

19.

a

b 808.1 ft

21.

a

b 68.2 km

23.

a

b 1.23 mi + 0.99 mi; 0.22 mi

25.

a

b 322.6 m

27.

a \(1^{\circ}\)

b \(66^{\circ}\)

c 2617.2 ft

d 1022.6 ft

29. 540,000 AU \(\approx 8.1 \times 10^{13}\) km

31. 750,000 AU \(\approx 1.1 \times 10^{14}\) km

33.

a \(\dfrac{3}{2}\)

b No, \(a\) is too short.

c 2

d 1

35.

a 1,

b 0,

c 2,

d 1,

37.

a \(C=25.37^{\circ}, B=114.63^{\circ}, b=16.97\)

b \(C=58.99^{\circ}, B=81.01^{\circ}, b=9.22\) or \(C=121.01^{\circ}, B=18.99^{\circ}, b=3.04\)

c no solution

d 5.14

39. \(A=40.44^{\circ}, B=114.56^{\circ} \text { or }A=139.56^{\circ}, B=15.44^{\circ}\)

41. \(C=37.14^{\circ}, A=93.86^{\circ}\)

43. 1299 yd or 277.2 yd

45.

a 11.79

b 24.16

c 24.15

47.

a \(\dfrac{1}{2} a b \sin C\)

b \(\dfrac{1}{2} a c \sin B\)

c \(\dfrac{1}{2} b c \sin A\)

49.

a

b \(b=\dfrac{h}{\sin A}\)

c \(h=a \sin B\)

d \(b=\dfrac{a \sin B}{\sin A}\)

e ii

3.3 The Law of Cosines

Homework 3.3

1

a \(74-70 \cos \theta\)

b \(12.78\)

c \(135.22\)

3.

a \(\dfrac{a^2+c^2-b^2}{2 a c}\)

b -0.4

5.

a \(b^2-8(\cos \alpha) b-65=0\)

b \(11.17,-5.82\)

7. 7.70

9. 13.44

11. 5.12

13. \(133.43^{\circ}\)

15. \(40.64^{\circ}\)

17. \(A=91.02^{\circ}, B=37.49^{\circ}, C=51.49^{\circ}\)

19. \(A=34.34^{\circ}, B=103.49^{\circ}, C=42.17^{\circ}\)

21. 6.30 or 2.70

23. 29.76 or 5.91

27. Law of Cosines: \(61^2=29^2+46^2-2 \cdot 29 \cdot 46 \cos \phi\)

29. Law of Sines: \(\dfrac{a}{\sin 46^{\circ}}=\dfrac{16}{\sin 25^{\circ}}\)

31. First the Law of Cosines: \(x^2=47^2+29^2-2 \cdot 47 \cdot 29 \cos 81^{\circ}\), then either the Law of Sines: \(\dfrac{\sin \theta}{47}=\dfrac{\sin 81^{\circ}}{x}\) or the Law of Cosines: \(47^2 = x^2 + 29^2 - 2 \cdot x \cdot 29 \cos \theta\)

33. Law of Cosines: \(9^2=4^2+z^2-2 \cdot 4 \cdot z \cos 28^{\circ}\) or use the Law of Sines first to find the (acute) angle opposite the side of length 4, then find the angle opposite the side of length \(z\) by subtracting the sum of the known angles from \(180^{\circ}\), then using the Law of Sines again.

35.

a

b \(b=16.87, A=85.53^{\circ}, C=47.47^{\circ}\)

37.

a

b \(A=58.41^{\circ}, B=48.19^{\circ}, C=73.40^{\circ}\)

39.

a

b \(a=116.52, A=85.07^{\circ}, C=56.93^{\circ}\) or
\(a=37.93, A=18.93^{\circ}, C=123.07^{\circ}\)

41.

a

b \(a=7.76, b=8.97, C=39^{\circ}\)

43.

a

b 1383.3 m

45.

a

b \(2123 \mathrm{mi}, 168.43^{\circ}\) east of north

46.

a

b \(7.74^{\circ}\) west of south, \(917.9 \mathrm{mi}\)

47.

a

b \(92.99 \mathrm{ft}\)

51. \(147.73 \mathrm{~cm}^2\)

53. 10.53

55. 4.08

57.

a First figure: \(b-x\) is the base of the small right triangle. Second: \(-x\) is the horizontal distance between \(P\) and the \(x\)-axis, so \(b+(-x)\) or \(b-x\) is the base of the large right triangle. Third: \(x=0\), and \(b\) is the base of a right triangle.

b First: \(x\) and \(y\) are the legs of a right triangle, \(a\) is the hypotenuse. Second: \(-x\) and \(y\) are the legs of a right triangle with hypotenuse \(a\). Third: \(x=0\) and \(y=a\)

c \(x=a \cos C\)

59.

\begin{aligned}
b^2 + c^2 &= (a^2 + c^2 - 2ac \cos B) + (a^2 + b^2 - 2bc \cos C) \\
&= 2a^2 + b^2 + c^2 - 2a (c \cos B + b \cos C)
\end{aligned}

so \(2a^2 = 2a(c \cos B + b \cos C\), and dividing both sides by \(2a\) yields \(a = (c \cos B + b \cos C)\).

61. For the first equation, start with the Law of Cosines in the form

\(a^2=b^2+c^2-2 b c \cos A\)

Add \(2 a b+2 b c \cos A-a^2\) to both sides of the equation, factor the right side, then divide both sides by \(2 b c\).

For the second equation, start with the Law of Cosines in the form

\(b^2+c^2-2 b c \cos A=a^2\)

Add \(2 b c-b^2-c^2\) to both sides of the equation, factor the right side, then divide both sides by \(2 b c\).

3.4 Chapter 3 Summary and Review

Chapter 3 Review Problems

1. \(\dfrac{1}{2}, \dfrac{\pm \sqrt{3}}{2}\)

3.

a

b 49.33

c

\(114^{\circ}\)

5.

a

b \(\cos \theta=\dfrac{-2}{\sqrt{13}}, \sin \theta=\dfrac{3}{\sqrt{13}}, \tan \theta=\dfrac{-3}{2}\)

c \(\theta=123.7^{\circ}\)

7.

a

b \(\cos \theta=\dfrac{-4}{5}, \sin \theta=\dfrac{3}{5}, \tan \theta=\dfrac{-3}{4}\)

c \(\theta=143.1^{\circ}\)

9.

a

b \(\cos \theta=\dfrac{-\sqrt{11}}{6}, \sin \theta=\dfrac{5}{6}, \quad \tan \theta=\dfrac{-5}{\sqrt{11}}\)

c \(\theta=123.6^{\circ}\)

11.

a

b \(\cos \theta=\dfrac{-7}{25}, \quad \sin \theta=\dfrac{24}{25}, \quad \tan \theta=\dfrac{-24}{7}\)

c \(\theta=106.3^{\circ}\)

13. \(9.9^{\circ}, 170.1^{\circ}\)

15. \(22.0^{\circ}, 158.0^{\circ}\)

17.

a \(7 \sqrt{2}\)

b \(28 \sqrt{2}\)

19. 5127.39 sq ft

21. \(20.41^{\circ}\)

23. \(a=27.86\)

25. \(b = 6.03\)

27. \(w = 62.10\)

29. \(s = 15.61 \text{ or } 57.45\)

31.

a

b 8.82

33.

a

b \(32.57^{\circ}\)

35.

a

b 16.29

37.

a

b \(58.65^{\circ}\)

39.

a

b 17.40

or

a

b 80.93

41.

a

b 16.08 mi, 80.4 mph

43.

a

b 72.47

45.

a

353.32

b 217.52 m

47.

a 79.64 m

b \(35.2^{\circ}\)

c 46.12 m

49. \(6.1^{\circ}\)

51.

4.2

53.

22.25 ft

55. 79,332.6 AU

57.

a \(O W\) bisects the central angle at \(O\), and the inscribed angle \(\theta\) is half the central angle at \(O\)

b \(\sin \theta=\dfrac{s}{2 r}\)

c \(r=\dfrac{s}{2 \sin \theta}\)

d \(d=\dfrac{s}{\sin \theta}\)