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Trigonometry : Law of Cosines

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Example Question #1 : Law Of Cosines

Which famous theorem does the Law of Cosines boil down to for right triangles?

Possible Answers:

Vertical Angle Theorem

Pythagorean Theorem

Isosceles Triangle Theorem

Cosine Theorem

Mean Value Theorem

Correct answer:

Pythagorean Theorem

Explanation:

The Law of Cosines is as follows:

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

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

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

Notice these equations contains the Pythagorean Theorem, $a^{2}+b^{2}= c^{2}$ , within it.

The term at the end is the adjusting term for triangles which are not right triangles.

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Example Question #2 : Law Of Cosines And Law Of Sines

Find the value of to the nearest tenth.

Possible Answers:

6.5

8.6

12.3

9.1

4.9

Correct answer:

8.6

Explanation:

This is a prime example of a case that calls for using the Law of Cosines, which states

$c^2=a^2+b^2-2ab \: cos(C)$

where , , and are the three sides of the triangle, and is the angle opposite side . Looking at our triangle, taking C=26^o , then we have a=15 , b=19 , and c=x . Plugging this into our formula, we get.

x^2=15^2+19^2-2(15)(19)cos(26^o)

Using our calculator to approximate the cosine value gives

x^2=225+361-2(15)(19)(0.8988)

Simplifying further gives

$x^2\approx 586-512.31=73.69$

Solving by taking the square root gives

$x\approx8.6$

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Example Question #1 : Law Of Cosines

Screen_shot_2015-03-07_at_1.33.25_pm

Using the Law of Cosines, determine the perimeter of the above triangle.

Possible Answers:

3.22

12.78

12.22

4.67

13.22

Correct answer:

12.22

Explanation:

To apply the Law of Cosines, is the unknown, and are the respective given sides, and the given angle is .

Therefore, the equation becomes:

c^2=(4^2)+(5^2)-2(4)(5)cos(40)=16+25-30.64=10.36

Which yields c=3.22

Add to the other two given sides to get the perimeter, 3.22+4+5=12.22

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Example Question #1 : Law Of Cosines

Solve for x:

Cosines 1

Possible Answers:

16.37

17.83

25.92

32.12

27.82

Correct answer:

16.37

Explanation:

We can solve for x using the law of cosines, $c^2 = a^ 2 + b ^ 2 - 2ab \cdot \cos C$ where C is the angle between sides a and b.

In this case:

$x^ 2 = 11^2 + 23^2 - 2(11)(23) \cdot \cos (41^o )$

x^2 = 650 - 506(0.7547 )

x^ 2 = 268.117

x = 16.37

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Example Question #5 : Law Of Cosines And Law Of Sines

Find the missing angles and sides.

Loc 6

Possible Answers:

None of the other answers.

$\\ a=7.2\:\textup{ft} \\ B=77^\circ \\ C=103^\circ$

$\\ a=6.1\:\textup{ft} \\ B=36^\circ \\ C=119^\circ$

$\\ a=5.4\:\textup{ft} \\ B=44.7^\circ \\ C=110.3^\circ$

$\\ a=5.4\:\textup{ft} \\ B=44.7^\circ \\ C=19.7^\circ$

Correct answer:

$\\ a=5.4\:\textup{ft} \\ B=44.7^\circ \\ C=110.3^\circ$

Explanation:

The Law of Cosines come in different forms depending on which angle or side you wish to find. One of the missing bits of information about our triangle is side length a. It is important to find this side because with side length a we can use the Law of Sines to easily find the angle measures. Side a "unlocks" the problem.

The pertinent LOC is $a^2=b^2+c^2-2bc\hspace{1mm}\cos\hspace{1mm}A$ .

$a^2=9^2+12^2-2(9)(12)\hspace{1mm}\cos\hspace{1mm}25=29.2375\rightarrow \sqrt{a^2}=\sqrt{29.3}=5.4072$

Now that we know side a, we can use the reciprocal form of the Law of Sines to find the remaining angle measures.

Angle B:

$\frac{\sin B}{b}=\frac{\sin A}{a}\rightarrow \sin B=b(\frac{\sin A}{a})=9(\frac{\sin 25^{\circ}}{5.4072})=0.7034$

To find the corresponding angle we take the inverse sine. $\sin^{-1}(0.7034)=44.7^{\circ}$

But there are two angles between 0° and 180°; there is 44.7° and $180^{\circ}-44.7^{\circ}=135.3^{\circ}$ . How do we know which angle to choose? We find out by solving for the last angle C with both of our hypothetical angles for angle B. Since side c is the largest side, it follows it should have the largest angle of all three angles in the triangle. Compute the measure of angle C by subtracting the given angle (angle A) and the angle we calculated (angle B) from 180°. Do this once with 44.7° and once with 135.3°. The first case results in the largest angle C and fits with c being the largest side. Thus angle B=44.7° and angle C must equal 110.3°.

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Example Question #6 : Law Of Cosines And Law Of Sines

In the triangle below, a=22 , b=17 , and c=15 . Find the measure of $\angle A$ to the nearest tenth.

Triangle abc

Possible Answers:

$50.5^\circ$

There is not enough information.

$42.9^\circ$

$86.6^\circ$

$90.0^\circ$

Correct answer:

$86.6^\circ$

Explanation:

To find an angle in an oblique triangle where all sides are known, use the law of cosines:

$cosA=\frac{-a^2+b^2+c^2}{2bc}$

$cosA=\frac{-22^2+17^2+15^2}{2(17)(15)}$

$A=cos^{-1}\left ( \frac{1}{17}\right )$

$A\approx86.6^\circ$

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Example Question #7 : Law Of Cosines And Law Of Sines

In the triangle below, $m\angle B=55^\circ$ , a=11 meters, and c=7 meters. What is the length of b, to the nearest tenth of a meter?

Triangle abc

Possible Answers:

There is not enough information.

13.0 meters

9.0 meters

8.5 meters

5.7 meters

Correct answer:

9.0 meters

Explanation:

The law of cosines states that $b^2=a^2+c^2-2ac\cos B$ .

So:

$b^2=11^2+7^2-2(11)(7)\cos 55^\circ$

$b\approx9.0$

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Example Question #1 : Law Of Cosines

A radar tower detects two ships. Ship A is 730 meters away and $20^\circ$ south of west. Ship B is 525 meters away and $45^\circ$ north of west. What is the distance between the two ships to the nearest meter?

Possible Answers:

507 meters

297 meters

516 meters

899 meters

696 meters

Correct answer:

696 meters

Explanation:

The sketch of the situation below shows that the angle between the ships from the radar station is 65 degrees.

Radar

To find the distance between the ships, use the law of cosines:

$c^2=a^2+b^2-2ab\cos C$

$c=\sqrt{525^2+730^2-2(525)(730)\cos65^\circ}$

$c\approx696$

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Example Question #9 : Law Of Cosines And Law Of Sines

In the triangle below, $m\angle C = 43.5^\circ$ , a=15.3 , and b=14.7 . What is the measure of $\angle A$ to the nearest tenth of a degree?

Triangle abc

Possible Answers:

$61.2^\circ$

$65.4^\circ$

$45.7^\circ$

$71.1^\circ$

There is not enough information.

Correct answer:

$71.1^\circ$

Explanation:

To find $\angle A$ , you must first find side c using the law of cosines:

$c^2=a^2+b^2-2ab\cos C$

$c=\sqrt{15.3^2+14.7^2-2(15.3)(14.7)\cos43.5^\circ}$

$c\approx 11.13$

Knowing c, you can find $\angle A$ using the law of sines or the law of cosines.

Law of sines:

$\frac{sinA}{a}=\frac{sinC}{c}$

$\frac{sinA}{15.3}=\frac{sin43.5^\circ}{11.13}$

$A=cos^{-1}\left ( \frac{15.3sin43.5^\circ}{11.13}\right )$

$A\approx71.1^\circ$

Law of cosines:

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

$A=cos^{-1}\left ( \frac{-15.3^2+14.7^2+11.13^2}{2(14.7)(11.13)}\right )$

$A\approx 71.1^\circ$

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Example Question #10 : Law Of Cosines And Law Of Sines

Given the triangle ABC , where a=8 , c=7 , and $\measuredangle B=135^{\circ}$ , calculate the side length to the thousandth decimal point.

Triangle abc

Possible Answers:

b=13.863

b=14.261

b=13.862

b=13.864

b=14.923

Correct answer:

b=13.863

Explanation:

Recall the law of cosines to determine the length of one side of a triangle ABC given the lengths of the other sides and and their included angle $\Theta$ :

$c^2=a^2+b^2-2ab(cos\Theta)$

Here, the unknown side length is denoted , and the other sides and the included angle is given. Substitute these values into the law of cosines and estimate square roots to the nearest thousandth decimal place to determine the side length .

$b^2=8^2+7^2-2(8)(7)(cos135^{\circ})$

$=113-112(-\frac{\sqrt(2)}{2})$

$\approx113+79.184 = 192.184$

$\Rightarrow b \approx 13.863$

Hence, the length of the remaining side of triangle ABC is approximately 13.863 units.

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Trigonometry : Law of Cosines

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Law Of Cosines

Example Question #2 : Law Of Cosines And Law Of Sines

Example Question #1 : Law Of Cosines

Example Question #1 : Law Of Cosines

Example Question #5 : Law Of Cosines And Law Of Sines

Example Question #6 : Law Of Cosines And Law Of Sines

Example Question #7 : Law Of Cosines And Law Of Sines

Example Question #1 : Law Of Cosines

Example Question #9 : Law Of Cosines And Law Of Sines

Example Question #10 : Law Of Cosines And Law Of Sines

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