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Precalculus : Law of Cosines and Sines

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Example Question #1 : Law Of Cosines And Sines

Which is NOT an angle of the following triangle?

(Not drawn to scale.)

Possible Answers:

$48.3^{\circ}$

$33.9^{\circ}$

$54.9^{\circ}$

$91.2^{\circ}$

Correct answer:

$48.3^{\circ}$

Explanation:

In order to solve this problem, we need to find the angles of the triangle. Only then will we be able to find which answer choice is NOT an angle. Using the Law of Cosines we are able to find each angle.

$\cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{9^{2}+11^{2}-6^{2}}{2\times 9\times 11}$

cos(A)=0.83

$A=\cos^{-1}(0.83)=33.9^{\circ}$

$\cos(B)=\frac{c^{2}+a^{2}-b^{2}}{2ac}=\frac{11^{2}+6^{2}-9^{2}}{2\times 6\times 11}$

cos(B)=0.575

$B=\cos^{-1}(0.575)=54.9^{\circ}$

To find angle we use the formula again or we can remember that the angles in a triangle add up to $180^{\circ}$ .

$C^{\circ}=180^{\circ}-33.9^{\circ}-54.9^{\circ}=91.2^{\circ}$

The answer choice that isn't an actual angle of the triangle is $48.3^{\circ}$ .

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Example Question #2 : Law Of Cosines And Sines

Solve the triangle

Law_of_cosines__sss_

Possible Answers:

$A = 99^{\circ}, B = 21^{\circ}, C=54^{\circ}$

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

None of the other answers

$A = 100^{\circ}, B = 31^{\circ}, C=49^{\circ}$

$A = 95^{\circ}, B = 23^{\circ}, C=62^{\circ}$

Correct answer:

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

Explanation:

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

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Example Question #3 : Use The Laws Of Cosines And Sines

Given three sides of the triangle below, determine the angles $\theta$ , $\phi$ , and $\beta$ in degrees.

Sss

Possible Answers:

$\theta \approx 45.0^\circ , \phi \approx 65.0^\circ , \beta \approx 70.0^\circ$

$\theta \approx 47.0^\circ , \phi \approx 74.9^\circ , \beta \approx 58.1^\circ$

$\theta \approx 75.5^\circ , \phi \approx 57.9^\circ , \beta \approx 46.6^\circ$

$\theta \approx 58.1^\circ , \phi \approx 47.0^\circ , \beta \approx 74.9^\circ$

Correct answer:

$\theta \approx 75.5^\circ , \phi \approx 57.9^\circ , \beta \approx 46.6^\circ$

Explanation:

We are only given sides, so we must use the Law of Cosines. The equation for the Law of Cosines is

$a^2=b^2+c^2-2bc\cos{A}$ ,

where , and are the sides of a triangle and the angle is opposite the side .

We have three known sides and three unknown angles, so we must write the Law three times, where each equation lets us solve for a different angle.

To solve for angle $\theta$ , we write

$16^2=12^2+14^2-2(12)(14) \cos{\theta}$ and solve for $\theta$ using the inverse cosine function $\arccos$ on a calculator to get

$\theta \approx 75.5$ .

Similarly, for angle $\phi$ ,

$14^2=12^2+16^2-2(12)(16) \cos{\phi}$

$\phi \approx 57.9$

and for $\beta$ ,

$12^2=14^2+16^2-2(14)(16) \cos{\beta}$

and $\beta \approx 46.6$

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Example Question #3 : Use The Laws Of Cosines And Sines

Lc2b

What is the measurement of $\angle C$ ? Round to the nearest tenth, if needed.

Possible Answers:

$83.8^\circ$

$83.7^\circ$

$83^\circ$

$84^\circ$

Correct answer:

$83.8^\circ$

Explanation:

We need to use the Law of Cosines for side then solve for cosC .

Therefore,

$\frac{c^2-a^2-b^2}{-2ab}=cosC$ .

Plugging in the information provided, we have:

$\frac{50^2-44^2-29^2}{-2(44)(29)}=cosC$ .

Then simplify, 0.1085423197=cosC .

To solve for , use $cos^{-1}(.1085423197)=C$ .

Solve and then round to the appropriate units: 83.76870649=83.8 . Therefore, $C=83.8^\circ$ .

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Example Question #4 : Use The Laws Of Cosines And Sines

Find the measure, in degrees, of the largest angle in a triangle whose sides measure , , and .

Possible Answers:

$102.8^\circ$

$119.8^\circ$

$111.8^\circ$

$87.8^\circ$

$93.8^\circ$

Correct answer:

$111.8^\circ$

Explanation:

When all three sides are given, Law of Cosines is appropriate.

Since 10 is the largest side length, it is opposite the largest angle and thus should be the c-value in the equation below.

$\\* c^2 = a^2 + b^2 - 2ab \cos (C) \\ \\* 10^2 = 5^2 + 7^2 - 2 \cdot 5 \cdot 7 \cos \(C) \\\\* C = \arccos \left(\frac {10^2-5^2-7^2}{2 \cdot 5 \cdot 7}\right) \approx 111.8^\circ$

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Example Question #1 : Use The Laws Of Cosines And Sines

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Possible Answers:

b=12

b=6.6

b=8

b=15.2

b=135

Correct answer:

b=6.6

Explanation:

We need to use the Law of Cosines in order to solve this problem

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

in this case, $a=6, c=10, B=40^{\circ}$

In order to arrive at our answer, we plug the numbers into our formula:

$b^{2}=6^{2}+10^{2}-2(6)(10)\cos (40^{\circ})$

$b^{2}=36+100-91.9$

$b\approx6.6$

Note: we use the "approximately" to indicate the answer is around 6.6. It will vary depending on your rounding.

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Example Question #3 : Law Of Cosines And Sines

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Possible Answers:

a=17

a=9.6

a=131.5

a=74

a=93

Correct answer:

a=9.6

Explanation:

In order to solve this problem, we need to use the following formula

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

in this case, $b=11, c=7, A=60^{\circ}$

We plug our numbers into our formula and get our answer:

$a^{2}=11^{2}+7^{2}-2(11)(7)\cos (60^{\circ})$

$a^{2}=121+49-77$

$a^{2}=93$

$a\approx 9.6$

Note: we use the "approximately" to indicate that the answer is around 9.6. It will vary depending on your rounding.

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Example Question #2 : Law Of Cosines And Sines

The 2 sides of a triangle have lengths of 10 and 20. The included angle is 25 degrees. What is the length of the third side to the nearest integer?

Possible Answers:

Correct answer:

Explanation:

Write the formula for the Law of Cosines.

c^2=a^2+b^2-2abcos(C)

Substitute the side lengths of the triangle and the included angle to find the third length.

c^2=10^2+20^2-2(10)(20)cos(25)

c^2=100+400-400cos(25)

c^2=500-400(.9063)

c^2=137.48

c=11.725

Round this to the nearest integer.

c=12

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Example Question #3 : Law Of Cosines And Sines

What is the approximate length of the unknown side of the triangle if two sides of the triangle are and , with an included angle of $30^\circ$ ?

Possible Answers:

12.4

13.4

20.5

12.9

Correct answer:

12.4

Explanation:

Write the formula for the Law of Cosines.

c^2=a^2+b^2-2abcos(C)

Substitute the known values and solve for .

c^2=10^2+20^2-2(10)(20)cos(30)

$c^2=100+400-400(\frac{\sqrt3}{2})$

$c^2=500-200\sqrt3$

$c=\sqrt{500-200\sqrt3} =12.3931=12.4$

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Example Question #6 : Law Of Cosines And Sines

Lc1c

What is the measurement of side using the Law of Cosines? Round to the nearest tenth.

Possible Answers:

234.2m

234m

234.3m

235m

Correct answer:

234.3m

Explanation:

The Law of Cosines for side is,

$a^2=b^2+c^2-2bc\cdot cosA$ .

Plugging in the information we know, the formula is,

a^2=400^2+375^2-2(400)(375)cos(35)=54879.38671 .

Then take the square of both sides: a=234.2634985 .

Finally, round to the appropriate units: a=234.3m .

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Precalculus : Law of Cosines and Sines

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