Solving Triangles and the Law of Cosines In this section we work out - - PowerPoint PPT Presentation

Elementary Functions

Part 5, Trigonometry Lecture 5.4a, The Law of Cosines

• Dr. Ken W. Smith

Sam Houston State University

2013

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Solving Triangles and the Law of Cosines

In this section we work out the law of cosines from our earlier identities and then practice applying this new identity. c2 = a2 + b2 − 2ab cos C. (1) Draw the triangle △ABC on the Cartesian plane with the vertex C at the

• rigin.

In the drawing sin C = y

b and cos C = x b . We may relabel the x and y

coordinates of A(x, y) as x = b cos C and y = b sin C.

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Solving Triangles and the Law of Cosines

We get information if we compute c2. By the Pythagorean theorem, c2 = (y2) + (a − x)2 = (b sin C)2 + (a − b cos C)2 = b2 sin2 C + a2 − 2ab cos C + b2 cos2 C. We use the Pythagorean identity to simplify b2 sin2 C + b2 cos2 C = b2 and so c2 = a2 + b2 − 2ab cos C

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One Angle and the Law of Cosines

c2 = a2 + b2 − 2ab cos C. It is straightforward to use the law of cosines when we know one angle and its two adjacent sides. This is the Side-Angle-Side (SAS) case, in which we may label the angle C and its two sides a and b and so we can solve for the side c. Or, if we have the Side-Side-Side (SSS) situation, in which we know all three sides, we can label one angle C and solve for that angle in terms of the sides a, b and c, using the law of cosines.

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A Worked Problem.

Solve the triangle C = 32◦, a = 100 feet, b = 150 feet.

• Solution. If C = 32◦, a = 100 feet, b = 150 feet then, by the law of

cosines (c2 = a2 + b2 − 2ab cos C), c2 = (100)2 + (150)2 − 2(100)(150) cos(32◦) ≈ 7059. So c ≈ √ 7059 ≈ 84.018 feet. Now we apply the law of sines: sin 32◦ 84.018 = sin A 100 = sin B 150 This forces sin A = 100 sin 32◦ 84.018 ≈ 0.6307 and sin B = 150 sin 32◦ 84.018 ≈ 0.9461 If the sine of A is 0.6307 then either A ≈ sin−1(0.63007) ≈ 39.1◦ or A ≈ 180 − 39.1 = 140.9◦.

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A Worked Problem.

We solve the triangle C = 32◦, a = 100 feet, b = 150 feet. We found A ≈ sin−1(0.63007) ≈ 39.1◦ or A ≈ 180 − 39.1 = 140.9◦. The second answer is too big, so A = 39.1◦. Since sin B = 0.9461 then either B ≈ sin−1(0.94510) ≈ 71.1◦ or B = 180 − 71.1 = 108.9◦. The second answer, B = 108.9◦ , makes perfect sense because the angles

• f the triangle need to add up to 180 degrees.

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The law of cosines and SSS

Solve the triangle a = 30 feet, b = 20 feet, c = 15 feet.

• Solution. If a = 30 feet, b = 20 feet, c = 15 feet then the law of cosines

tells us that 152 − 302 − 202 −2(30)(20) = cos C so −1075 −1200 = 43 48 = cos C Therefore C = cos−1(43 48) ≈ 26.38◦ . By the law of sines sin 26.38◦ 15 = sin A 30 = sin B 20 , so sin A = 2 sin 26.38◦. and so A = 62.7◦ or A = 180◦ − 62.7◦ = 117.3◦

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The law of cosines and SSS

We solve the triangle a = 30 feet, b = 20 feet, c = 15 feet. We found A = 62.7◦ or A = 180◦ − 62.7◦ = 117.3◦ Also sin B = 4 3 sin 26.38◦ so B = 36.33◦ or 180◦ − 36.33◦ = 143.67◦ The second result here is too large. Our final answer, after checking that angles add up to 180◦, is A = 117.3◦, B = 36.3◦, C = 26.4◦.

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Some Worked Problems.

A triangle has angle C = 20◦ and sides a = 10, b = 20. Use the law of cosines to find the length of the side c. Solution. c2 = 102 + 202 − 2(10)(20) cos 20◦ = 500 − 400 cos 20◦ ≈ 124.1 So c ≈ √ 124.1 ≈ 11.14. Use the law of sines to find two possible values of the angle A (one in which A is acute and one in which A is obtuse.)

• Solution. The sine of A is 0.307 so A = 17.88◦ or 162.12◦.

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Some Worked Problems.

A triangle has angle C = 20◦ and sides a = 10, b = 20. Use the law of sines to find two possible values of the angle B (one in which B is acute and one in which B is obtuse.)

• Solution. The sine of B is 0.614 so B ≈ 37.88◦ or 142.12◦.

Solve the triangle.

• Solution. c ≈ 11.14, A ≈ 17.88◦ and B ≈ 142.12◦.

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Solving Triangles and the Law of Cosines

Solve the triangle with sides a = 15, b = 12, c = 10.

• Solution. By the Law of Cosines, C = 41.65◦. Now use the Law of Sines

to get a = 15, b = 12, c = 10, A = 85.46◦, B = 52.89◦, C = 41.65◦.

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Solving Triangles and the Law of Cosines

Solve the triangles with sides a = 15, b = 12, c = 30.

• Solution. Since one side is longer than the sum of the other two sides,

no such triangle is possible. (No calculations are necessary here. If we did a calculation, we would eventually take the arccosine of a number larger than 1, which is impossible.)

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The Law of Cosines

In the next presentation, we will look at Heron’s formula and the “Five Guys”, (five trig identities I give to my classes.) (End)

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Elementary Functions

Part 5, Trigonometry Lecture 5.4b, Heron’s Formula, Five Guys

• Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 14 / 22

The area of an oblique triangle

Since a triangle is half of a parallelogram, its area is one-half of the product of its base and height. Let K represent the area of a triangle. Looking at the drawing below, we see that K = 1

2ch.

But earlier, in our proof of the Law of Sines, we solved for h and we wrote h = b sin A and h = a sin B. So we can substitute for h and write the area as K = 1

2cb sin A and K = 1 2ac sin B

Or we could call the known angle C and just write K = 1 2ab sin C (2)

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The area of an oblique triangle

K = 1 2ab sin C (3) We can summarize this by saying that the area of a triangle is one-half of the product of the sine of an angle and its neighboring sides.

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Heron’s formula

If we know the three sides a, b and c then in theory, since the triangle is fixed and we can compute the three angles, we should be able to compute the area of the triangle. A first step is the formula we found when we proved the Law of Sines: K = 1 2ab sin C A succinct formula for the area of a triangle, given the three sides, was worked out long ago by Heron of Alexandria.

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Heron’s formula

Let’s take the area equation K = 1

2ab sin C

and square both sides. K2 = 1

4a2b2 sin2 C.

By the Pythagorean identity replace sin2 C by 1 − cos2 C K2 = 1

4a2b2(1 − cos2 C).

and distribute K2 = 1

4a2b2 − 1 4a2b2 cos2 C.

Now use the Law of Cosines in the form cos C = a2+b2−c2

2ab

to replace cos2 C: K2 = 1

4a2b2 − ( 1 4a2b2)( a2+b2−c2 2ab

)2 and simplify K2 = 1

4a2b2 − ( 1 16)(a2 + b2 − c2)2

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Heron’s formula

Using the law of sine and the law of cosines, we worked out a formula for the area: K2 = 1

4a2b2 − ( 1 16)(a2 + b2 − c2)2

K2 = 1

16(4a2b2 − (a2 + b2 − c2)2)

Now, with a little bit of algebra ... (skipping a bunch of steps!) ... we can get this into the form K2 = ( a+b+c

2

)( −a+b+c

2

)( a−b+c

2

)( a+b−c

2

) The perimeter of the triangle is a + b + c. Half of the perimeter, written s = a+b+c

2

is called the semiperimeter. We can use the semiperimeter s to shorten this equation K2 = s(s − a)(s − b)(s − c). So the area of a triangle is K =

• s(s − a)(s − b)(s − c).

This is Heron’s formula

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Summary & Five Guys

Summary of our identities We have come across two very useful identities that are easy to remember:

1 The Pythagorean identity, cos2 θ + sin2 θ = 1 (and its two siblings),

and

2 The Law of Sines, sin A a

= sin B

b

= sin C

c .

There are five more identities that are very useful and if we know them (or have them handy) then the other trig identities follow easily from them. Our goal here is understanding, not memorization! But if one were to memorize trig identities, in addition to the easy two above, I recommend the following five.

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Five Guys

1 The formulas for the sine and cosine of the sum of two angles:

cos(α + β) = cos α cos β − sin α sin β sin(α + β) = sin α cos β + cos α sin β

2 The power reduction formulas:

cos2 α = 1+cos(2α)

2

sin2 α = 1−cos(2α)

2 3 The law of cosines:

c2 = a2 + b2 − 2ab cos C In my precalculus and trigonometry classes these five identities, boxed above, will be provided on quizzes and exams. Since I have never memorized these – and I’ve done well in mathematics, emphasizing understanding over memorization – then I won’t ask my students to memorize them. Students should focus on understanding how to wield these five guys in a variety of environments.

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Five Guys

In the next presentation, we will look at polar coordinates. (End)

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