There are two more trigonometric functions that will concern us in physics. The sine of an angle is the opposite side divided by the hypotenuse, while the cosine of an angle is the adjacent side divided by the hypotenuse.

8. Find the sine and cosine of each angle (abbreviated as sin and cos)

a)	b)	c)
a) sin(Ø) = opposite/hypotenuse sin(Ø) = (18.6 cm)/(39.8 cm) sin(Ø) = .47 cos(Ø) = adjacent/hypotenuse cos(Ø) = (35.2 cm)/(39.8 cm) cos(Ø) = .88	b) sin(Ø) = opp/hyp sin(Ø) = (3.15 ft)/(3.45 ft) sin(Ø) = .91 cos(Ø) = adj/hyp cos(Ø) = (1.4 ft)/(3.45 ft) cos(Ø) = .41	c) sin(Ø) = opp/hyp sin(Ø) = (660 cm)/(30.0 ft) sin(Ø) = (660 cm)/[(30.0 ft)(12 in/ft)(2.54 cm/in)] sin(Ø) = .72 cos(Ø) = adj/hyp cos(Ø) = (251 in)/(30.0 ft) cos(Ø) = (251 in)/[(30.0 ft)(12 in/ft)] cos(Ø) = .70

Although the tangent can take on any value from zero to infinity, the sine and cosine are restricted by their definitions to the range zero to one. This is because the hypotenuse (and hence the denominator of the fraction) is always larger than either leg.

Negative trigonometric functions occur, but will be left to your trig class.

9. Use the sine to find the missing side or angle:

a)	b)	c)
a) sin(Ø) = opp/hyp sin(Ø) = (32.6 cm)/(60.1 cm) sin(Ø) = .542 Ø = 32.8°	b) sin(Ø) = opp/hyp sin(52.6°) = x/(87.5 ft) x = (87.5 ft)(sin 52.6) x = 69.5 ft	c) sin(73.1°) = (4.68 m)/x x = (4.68 m)/(sin 73.1) x = 4.89 m

10. Use the cosine to find the missing side or angle:

a)	b)	c)
a) cos(Ø) = adj/hyp cos(Ø) = (31.0 ft)/(33.6 ft) cos(Ø) = .923 Ø = 22.7°	b) cos(Ø) = adj/hyp cos(71.6°) = x/(11.7 m) x = (11.7 m)(cos 71.6) x = 3.69 m	c) cos(Ø) = adj/hyp cos(58.4°) = (289 cm)/x x = (289 cm)/(cos 58.4) x = 551 cm

With these three functions you can solve any right triangle. The only thing that is tricky for the novice is determining which trig function to use. Here is the step-by-step approach that will get you a solution:

1) Identify the right angle.
2) Identify the acute angle and the two sides to be considered. (One of them will be your unknown.)
3) Determine the appropriate trig function (depending on whether you have opposite and adjacent, opposite and hypotenuse, etc. for your sides).
4) Write out the definition of the function (e.g. tanØ = opposite/adjacent).
5) Plug in your two known values.
6) Solve.

The key to this is being systematic in your approach, and being confident that the solution will fall into your lap. Here's your big chance:

11. Find the missing side or angle by use of the appropriate function.

a)	b)	c)
d)	e)	f)

a) tan(Ø) = opp/adj tan(31.6°) = x/(16.7 ft) x = (16.7 ft)(tan 31.6) x = 10.3 ft	b) sin(Ø) = (33.2 cm)/(36.5 cm) sin(Ø) = .910 Ø = 65.4°	c) sin(Ø) = opp/hyp sin(Ø) = a/x x = a/sinØ
d) tan(Ø) = opp/adj tan(Ø) = [(7.32 ft)(12 in/1 ft)(2.54 cm/1 in)]/[(2.10 m)(100 cm/1 m)] tan(Ø) = 1.062 Ø = 46.7°	e) sin(58.2°) = x/(219 in) x = (219 in)(sin 58.2) x = 186 in	f) cos(Ø) = b/a Ø = cos-1(b/a)

Remember, if you know one side of a rectangle, the opposite side is equal. I will use this principle in the future, assuming you will not need to be told.

12*. Find the missing parts of the rectangles shown.

a)	b)	c)
a) cos(Ø) = adj/hyp cos(24.4°) = x/(19.2 cm) x = (19.2 cm)(cos 24.4) x = 17.5 cm sin(Ø) = opp/hyp sin(24.4°) = y/(19.2 cm) y = (19.2 cm)(sin 24.4) y = 7.93 cm	b) tan(Ø) = opp/adj tan(Ø) = (28.1 ft)/(23.3 ft) tan(Ø) = 1.21 Ø = 50.5° cos(Ø) = adj/hyp cos(50.5°) = (23.3 ft)/R R = (23.3 ft)/(cos 50.5) R = 36.6 ft note: you could have used sin(50.5°) = (28.1 ft)/R	c) cos(Ø) = adj/hyp cos(Ø) = [(8.35 in)(2.54 cm/in)]/(47.3 cm) cos(Ø) = .448 Ø = 63.4° sin(Ø) = opp/hyp sin(63.4°) = y/(47.3 cm) y = (47.3 cm)(sin 63.4) y = 42.3 cm

13**. Find the unknown angle:

sin(52°) = (31.1 cm)/a a = (31.1 cm)/(sin 52) a = 39.5 a2 + b2 = c2 b2 = c2 - a2 b2 = 46.32 - 39.52 b = 24.2 cm cos(Ø) = [(6.1 in)(2.54 cm/in)]/(24.2 cm) cos(Ø) = .640 Ø = 50.2°

14. Write expressions for the perimeter and area of each figure:

a)	b)	c)
a) P = 25 + 25tan32 + [252 + (25tan32)2]1/2 P = 25[1 + tan32 + (1 + tan232)1/2] P = 70.1 cm A = (1/2)bh A = (1/2)(25)(25tan32) A = 145 cm2	b) P = a + atanØ + [a2 + (atanØ)2]1/2 P = a[1 + tanØ + (1 + tan2Ø)1/2] A = (1/2)(a)(atanØ) (a2/2)/tanØ)	c) P = 17(1 + sin41 + cos41) P = 41.0 ft A = (1/2)(17cos41)(17sin41) A = 71.5 ft2

15*. The areas of the semi-circles are A1 and A2. Express q in terms of these areas.

T1 and T2 are radii of the semi-circles. Thus: A1 = 1/2(3.14)r12 r1 = [(2A1)/(3.14)]1/2 A2 = 1/2(3.14)r22 r2 = [(2A2)/(3.14)]1/2 tan(Ø) = 2r1/2r2 tan(Ø) = r1/r2 tan(Ø) = [(2A1)/(3.14)]1/2/[(2A2)/(3.14)]1/2 tan(Ø) = (A1/A2)1/2 Ø = tan-1(A1/A2)1/2