]> Chapter 07: Applications of Trigonometric Functions (Sullivan & Sullivan 3e)

Chapter 07: Applications of Trigonometric Functions

The majority of this chapter actually deals with solving triangles—finding the lengths of all three sides and the measures of all three angles. To that end, let me begin with the standard diagram for a generic triangle:

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We use the first three lowercase Roman letters to represent the lengths of the three sides, and the first three lowercase Greek letters to represent the measures of the three angles. One can also use uppercase Roman letters for the three points…but that’s rarely of interest.

Note that in order to solve a triangle, one must be given three pieces of information (out of the six that we ultimately could have). The list of possible cases is:

Three sides (SSS)
Two sides and an included angle (SAS)
Two sides and a non-included angle (SSA)
Two angles and an included side (ASA)
Two angles and a non-included side (AAS)
Three angles (AAA)

The AAA case cannot be solved—there are an infinite number of solutions.

For this chapter, we will pretty much use degrees exclusively…solving triangles is intimately related to real-life problems, and (alas) no one solves real world problems with radians.

A word of advice: where possible, stick to using the given (exact) information, rather than the derived (solved) information…that will make your answers more accurate.

§7.1: Right Triangle Trigonometry

Solving right triangles is very easy…you should already be able to do it.

A note on terminology: an angle of elevation is an angle measured up from the horizon. An angle of depression is an angle measured down from the horizon.

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Examples

[1.] Solve the triangle shown below.

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The other acute angle has measure $90 ° - 28 ° = 62 °$ .

For the missing leg: $\tan (28 °) = \frac{40}{L} \Rightarrow L = \frac{40}{\tan (28 °)} = 75.229$

For the hypotenuse: $\sin (28 °) = \frac{40}{h} \Rightarrow h = \frac{40}{\sin (28 °)} = 85.202$

[2.] A 20 foot long ladder is leaned against a house so that it makes a 70° angle with the ground. How far up the house does it reach? How far away from the house is the base of the ladder?

Here’s a diagram:

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For the height: $\sin (70 °) = \frac{h}{20} \Rightarrow h = 20 \sin (70 °) = 18.79$ feet.

For the base: $\cos (70 °) = \frac{b}{20} \Rightarrow b = 20 \cos (70 °) = 6.84$ feet.

§7.2: The Law of Sines

$\frac{\sin (α)}{a} = \frac{\sin (β)}{b} = \frac{\sin (γ)}{c}$

You’ll only use two pieces of this compound equation at a time…

Two Angles: AAS and ASA

When given two angles and one side, the Law of Sines can be used to solve the triangle. Finding the third angle is no problem at all…the Law of Sines will be used to find the lengths of the other two sides.

Two Sides: SSA

When given one angle and two sides, things can get dicey…there may be zero, one of two solutions! We will only consider the case with a non-included angle at this time (see the diagram).

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The problem is hinted at by the diagram: the side opposite the given angle may not be long enough to even make a triangle! The smallest possible length for the side opposite will be the length of the perpendicular.

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Of course, since that’s perpendicular, you can use good old Right Triangle Trigonometry to find that minimum length. If the side opposite is too short, then there is no solution for the given information.

The really interesting case is when the side opposite is long enough (longer than the perpendicular) but shorter than the side adjacent…because maybe the side opposite could sit at this angle…

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…or maybe it could sit at this other angle:

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When the side opposite is longer than the perpendicular, but shorter than the side adjacent, then there are two solutions.

The Law of Sines will only give you one of the solutions directly…you’ll have to use a bit of Geometry to find the other solution!

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This diagram shows both positions for the side opposite…note how together, they make an isosceles triangle. Thus, the base angles are congruent…but only the right-hand one is part of the triangle made by the side opposite. The angle made by the other solution is marked with the question mark. How are theta and the question mark related?

The next possibility is if the side opposite is longer than the side adjacent:

An Image

If the side opposite is longer than the side adjacent, then there is one solution.

Alas, all of these pictures make one small assumption: that the initially given angle is acute. If the given angle is non-acute (right or obtuse), then there is one solution only if the side opposite is longer than the side adjacent; otherwise there is no solution.

Examples

[3.] Solve the triangle with $α = 45 °$ , $β = 60 °$ and $a = 14$ .

That makes $γ = 180 ° - 45 ° - 60 ° = 75 °$ . Now for the sides:

$\frac{\sin (45 °)}{14} = \frac{\sin (60 °)}{b} \Rightarrow b = \frac{14 \sin (60 °)}{\sin (45 °)} = 17.146$

$\frac{\sin (45 °)}{14} = \frac{\sin (75 °)}{c} \Rightarrow c = \frac{14 \sin (75 °)}{\sin (45 °)} = 19.124$

Notice that I didn’t use the less accurate value for b to solve for c…instead, I used the more accurate value for a.

[4.] Solve the triangle with $a = 10$ , $b = 4$ and $β = 36 °$ .

This is SSA…I need to see how long the perpendicular is.

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$\sin (β) = \frac{L}{a} \Rightarrow L = a \sin (β) = 10 \sin (36 °) = 5.88$ ; since the side opposite is shorter than this, there is no solution.

[5.] Solve the triangle with $b = 3$ , $c = 2$ and $γ = 25 °$

And again!

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$\sin (25 °) = \frac{L}{3} \Rightarrow L = 3 \sin (25 °) = 1.27$ ; since the side opposite is longer than this, but shorter than the side adjacent, there are two solutions. I’ll have to use the Law of Sines to solve for one of the possible values of $β$ …and the nature of the inverse sine function means that I’ll find the acute solution first.

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$\frac{\sin (25 °)}{2} = \frac{\sin (β_{1})}{3} \Rightarrow \sin (β_{1}) = \frac{3 \sin (25 °)}{2} \Rightarrow β_{1} = \sin^{- 1} (\frac{3 \sin (25 °)}{2}) = 39.34 °$

$α_{1} = 180 ° - 25 ° - 39.34 ° =115.66 °$

$\frac{\sin (25 °)}{2} = \frac{\sin (115.66 °)}{a_{1}} \Rightarrow a_{1} = \frac{2 \sin (115.66 °)}{\sin (25 °)} = 4.266$

That first angle that I found is the key to the second solution…in particular, $β_{2}$ is the supplement of $β_{1}$ . Thus, $β_{2} = 180 ° - 39.34 ° = 140.66 °$

$α_{2} = 180 ° - 25 ° - 140.66 ° =14.34 °$

$\frac{\sin (25 °)}{2} = \frac{\sin (14.34 °)}{a_{2}} \Rightarrow a_{2} = \frac{2 \sin (14.34 °)}{\sin (25 °)} = 1.172$

Whew!

[6.] Solve the triangle with $a = 5$ , $c = 7$ and $γ = 112 °$ .

Another side-side-angle…since the side opposite is longer than the side adjacent, there is one solution.

$\frac{\sin (112 °)}{7} = \frac{\sin (α)}{5} \Rightarrow \sin (α) = \frac{5 \sin (112 °)}{7} = 0.662 \Rightarrow α {=sin}^{- 1} (0.662) = 41.474 °$

$β = 180 ° - 112 ° - 41.474 ° =26.527 °$

$\frac{\sin (112 °)}{7} = \frac{\sin (26.527 °)}{b} \Rightarrow b = \frac{7 \sin (26.527 °)}{\sin (112 °)} = 3.372$

[7.] A river flows from West to East. From two points 180m apart on the South side of the river, a surveyor sights a large boulder. From the West point, the boulder is at an angle 27° East of North. From the East point, the boulder is at an angle of 43° West of North. How wide is the river?

Here’s a picture of the situation.

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The apex angle of the triangles measures 70°; I’ll use that to solve for the East side.

$\frac{\sin (70 °)}{180} = \frac{\sin (63 °)}{e} \Rightarrow e = \frac{180 \sin (63 °)}{\sin (70 °)} = 170.674$

Now, notice that there is a little right triangle in there…and the distance that I want to find is one leg of that right triangle.

$\sin (47 °) = \frac{d}{170.674} \Rightarrow d = 170.674sin (47 °) = 124.823$

The distance across the river is 124.823 meters.

§7.3: The Law of Cosines

The Rules

$a^{2} = b^{2} + c^{2} - 2 b c \cos (α)$

$b^{2} = a^{2} + c^{2} - 2 a c \cos (β)$

$c^{2} = a^{2} + b^{2} - 2 a b \cos (γ)$

This is the more general form of the Pythagorean Theorem.

Three Sides: SSS

There may be no solution when three sides are specified. There will be one solution only if the triangle obeys the Triangle Inequality: the sum of any two side lengths is greater than the length of the third side.

Two Sides: SAS

There is one solution when two sides and the included angle are given.

Triangle Solving Summary

When Given…	Start solving with…	# of Solutions
SSS	Law of Cosines	0 or 1
SAS	Law of Cosines	1
SSA	Law of Sines	0, 1 or 2
AAS	Law of Sines	1
ASA	Law of Sines	1
AAA	None	Infinite

When starting with the Law of Sines, you don’t have to finish with it! In fact, I recommend that you switch and use the Law of Cosines as soon as possible, since there is no possibility of two solutions when using the Law of Cosines (unlike the Law of Sines!).

Examples

[8.] Solve the triangle with $a = 9$ , $b = 40$ and $c = 41$ .

This obeys the Triangle Inequality, and will have one solution.

$9^{2} = 40^{2} + 41^{2} - 2 \cdot 40 \cdot 41 \cdot \cos (α) \Rightarrow \cos (α) = \frac{9^{2} - 40^{2} - 41^{2}}{- 2 \cdot 40 \cdot 41} = 0.976 \Rightarrow α = \cos^{- 1} (0.976) = 12.68 °$

$40^{2} = 9^{2} + 41^{2} - 2 \cdot 9 \cdot 41 \cdot \cos (β) \Rightarrow \cos (β) = \frac{40^{2} - 9^{2} - 41^{2}}{- 2 \cdot 9 \cdot 41} = 0.2195 \Rightarrow β = \cos^{- 1} (0.2195) = 77.32 °$

$γ = 180 ° - 12.68 ° - 77.32 ° = 90 °$

Well…if I had known it was a right triangle, I could have saved myself a bit of work!

[9.] Solve the triangle with $a = 6$ , $b = 10$ and $c = 3$ .

Since $a + c < b$ , there is no solution.

[10.] Solve the triangle with $a = 16$ , $c = 14$ and $β = 120 °$ .

$b^{2} = 16^{2} + 14^{2} - 2 \cdot 16 \cdot 14 \cdot \cos (120 °) = 676 \Rightarrow b = 26$

$16^{2} = 26^{2} + 14^{2} - 2 \cdot 26 \cdot 14 \cdot \cos (α) \Rightarrow \cos (α) = \frac{16^{2} - 26^{2} - 14^{2}}{- 2 \cdot 26 \cdot 14} = 0.846 \Rightarrow α = \cos^{- 1} (0.846) = 32.204 °$

$γ = 180 ° - 120 ° - 32.204 ° = 27.796 °$

[11.] A surveyor stands at a reference point and sights the edges of a wooded lot. The left edge is at an angle 25° West of North, and at a distance of 15m. The right edge is at an angle 37° East of North, and at a distance of 28m. How wide is the wooded lot?

Here’s a picture of the situation:

An Image

$x^{2} = 15^{2} + 28^{2} - 2 \cdot 15 \cdot 28 \cdot \cos (62 °) = 614.644 \Rightarrow x = 24.792$ meters.

§7.4: The Area of a Triangle

There are more ways to find the area of a triangle than the old half base times height method…

The SAS Formula

$\frac{1}{2} b c \sin (α)$

$\frac{1}{2} a c \sin (β)$

$\frac{1}{2} a b \sin (γ)$

Note that these are really half base times height…just in disguise…

The SSS Formula

…also known as Heron’s Formula. Let s be the semi-perimeter: $s = \frac{1}{2} (a + b + c)$ . That makes the area of the triangle $\sqrt{s (s - a) (s - b) (s - c)}$ .

Examples

[12.] Find the area of the triangle with $a = 4$ , $b = 5$ and $γ = 150 °$ .

The area is $\frac{1}{2} (4) (5) \sin (150 °) = 5$ square units.

[13.] Find the area of the triangle with side lengths 10, 20 and 15.

The semi-perimeter is $\frac{1}{2} (10 + 20 + 15) = 22.5$ ; the area is $\sqrt{22.5 (22.5 - 10) (22.5 - 20) (22.5 - 15)} = 72.618$ square units.

§7.5: Harmonic Motion

We’ve already learned how to find the equation of simple periodic phenomena...what we learned applies to simple harmonic motion. The only new thing that we will add is that sometimes we will use cosine instead of sine…in particular, if the motion starts at an extreme (maximum or minimum) then we will use cosine—otherwise, we’ll still use sine.

Examples

[14.] A spring is stretched to 2cm. When released, it takes 0.6 seconds to reach full compression. Write the equation for this simple harmonic motion.

The amplitude is 2. If it takes 0.6 seconds to go from extension to compression, then it will take 1.2 seconds to go from full extension to compression and back to extension—thus, the period is 1.2 seconds. Since the period is $\frac{2 π}{b}$ , $\frac{2 π}{b} = 1.2 \Rightarrow \frac{2 π}{1.2} = b \approx 5.236$ . Since the motion begins with full extension (a maximum), I need to use cosine.

The equation of motion is $2 \cos (5.236 t)$ .

Page last validated 2010-08-15