trigonometric ratios assignment edgenuity answers

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Trigonometry

Course: trigonometry   >   unit 1.

• Reciprocal trig ratios
• Finding reciprocal trig ratios
• Using reciprocal trig ratios
• Sine & cosine of complementary angles

Trigonometric ratios review

What are the trigonometric ratios, practice set 1: sine, cosine, and tangent.

• Your answer should be
• a proper fraction, like 1 / 2 ‍   or 6 / 10 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• an improper fraction, like 10 / 7 ‍   or 14 / 8 ‍  
• a simplified improper fraction, like 7 / 4 ‍  

Practice set 2: cotangent, secant, and cosecant

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11.1: The Trigonometric Ratios

• Last updated
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• Page ID 122907

• Katherine Yoshiwara
• Los Angeles Pierce College

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. \(0 < x < 16\)

19. \(6\sqrt{2}\)in

21. \(w = 6\sqrt{10}\)in

25. \(\sqrt{3}\)

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.

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

39. \(8 \sqrt{3}\) in

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 \(4\sqrt{13} \approx 14.42\)

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 \(\sin \theta = 0.2116, \cos \theta = 0.9774, \tan \theta = 0.2165\)

(Answers may vary)

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

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

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

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. \(12 \sin 40^{\circ} \approx 7.71\)

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.

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 \(\cos \theta\).

a \(0.2358\)

c \(48^{\circ}\)

d \(77^{\circ}\)

a \(\dfrac{5}{12}\)

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.

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

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

2.3 Solving Right Triangles

Homework 2.3.

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

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

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

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

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

• 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 \(A\).

• 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 \(B\).

• 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. \(11.5^{\circ}\)

25. \(56.3^{\circ}\)

27. \(73.5^{\circ}\)

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 \(\sin 6^{\circ}\).

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

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

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

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

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

49. (a) and (b)

51. (a) and (d)

53. \(\dfrac{\sqrt{3}}{2} \approx 0.8660\)

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

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})

a \(32 \sqrt{3} \mathrm{~cm}\)

b \(128 \sqrt{3} \mathrm{sq} \mathrm{cm}\)

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 \(A+B+C>180^{\circ}\)

9. \(a = 97\)

11. \(c = 52\)

15. \(\theta=35.26^{\circ}\)

17. No. \(a=6, c=10\) or \(a=9, c=15\)

a \(w=86.05\)

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

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

35. 43.30 cm

37. 15.92 m

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

a \(60.26^{\circ}\)

b \(60.26^{\circ}\)

c \(m=\dfrac{7}{4}=\tan \theta\)

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

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

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

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Which letter represents the HIPPOTENUSE?

Which letter represents the ADJACENT cathetus to angle a?

Which letter represents the OPPOSITE cathetus to angle a?

Find   cos ⁡   a = \cos\ a= cos   a =    

  5 13 \frac{5}{13} 1 3 5 ​    

  7 13 \frac{7}{13} 1 3 7 ​    

  5 7 \frac{5}{7} 7 5 ​    

  13 7 \frac{13}{7} 7 1 3 ​    

  13 5 \frac{13}{5} 5 1 3 ​    

Find   tan ⁡   a = \tan\ a= tan   a =    

Find   sin ⁡   θ = \sin\ \theta= sin   θ =    

  6 10 \frac{6}{10} 1 0 6 ​    

  6 8 \frac{6}{8} 8 6 ​    

  6 12 \frac{6}{12} 1 2 6 ​    

  6 9 \frac{6}{9} 9 6 ​    

  6 11 \frac{6}{11} 1 1 6 ​    

If sin ⁡ θ = 12 15 \sin\ \theta\ =\ \frac{12}{15} sin θ = 1 5 1 2 ​ , what's the length of the OPPOSITE cathetus?

If   tan ⁡   θ   =   12 9 \tan\ \theta\ =\ \frac{12}{9} tan   θ   =   9 1 2 ​    , what's the length of the HYPOTENUSE?

Find the value of x

If    θ = 27 ° \theta=27° θ = 2 7 °   and the adjacent cathetus is equal to 4, find the approximate value of the OPPOSITE cathetus?

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