Law of Cosines - Trigonometry

Card 0 of 132

Question

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

Answer

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

Therefore, the equation becomes:

Which yields

Add to the other two given sides to get the perimeter,

Compare your answer with the correct one above

Question

In triangle $\triangle ABC$ , $\measuredangle A = 115^{\circ}$ , and . To the nearest tenth, what is ?

Answer

By the Law of Cosines,

$BC^2 = AB^2 +AC^2 - 2\left (AB\right )\left ( AC\right )\cos\angle A$

or, equivalently,

$BC = \sqrt[]{AB^2 +AC^2 - 2\left ( AB\right )\left ( AC\right )\cos \angle A}$

Substitute:

$BC = \sqrt[]{31^2 + 42^2 - 2\left ( 31\right )\left ( 42\right )\cos \left ( 115^{\circ}\right )}$

$\textup{BC}= \sqrt[]{961 + 1764 - 2604\cdot (-0.4226)} \approx \sqrt{2725 + 1100} \approx \sqrt{3825}\approx61.9$

Compare your answer with the correct one above

Question

A triangle has side lengths , and $\angle C=150^{\circ}$

Which of the following equations can be used to find the length of side ?

Answer

You are given the length of two sides of a triangle and the angle between them; therefore, you should use the Law of Cosines to find $\angle A$ , $\angle B$ , or, in this case, the length of .

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

Substitute the given values for , , and $\angle C$ :

$c^{2}=(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}$

At this point, if you are solving for , take the square root of both sides of the equation.

$\sqrt{c^{2}}=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

$c=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

This question merely asks for the equation, rather than the solution, so you need not simplify any further.

Two of the answer choices are equations derived from the Law of Sines. To use the Law of Sines, you must know at least one side and angle that correspond to one another, which is not the case here.

Compare your answer with the correct one above

Question

Given , and determine to the nearest degree the measure of $\angle B$ .

Answer

We are given three sides and our desire is to find an angle, this means we must utilize the Law of Cosines. Since the angle desired is $\angle B$ the equation must be rewritten as such:

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

Substituting the given values:

$5.1^{2} = 6.2^{2} + 3.5^{2} - 2\left(6.2 \right )\left(3.5 \right )\cos B$

Rearranging:

$\cos B = \frac{5.1^{2} - 3.5^{2} - 6.2^{2}}{-2\left(3.5 \right )\left(6.2)}$

Solving the right hand side and taking the inverse cosine we obtain:

$B = 55^{\circ}$

Compare your answer with the correct one above

Question

If , and , determine the measure of $\angle C$ to the nearest degree.

Answer

This is a straightforward Law of Cosines problem since we are given three sides and desire one of the corresponding angles in the triangle. We write down the Law of Cosines to start:

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

Substituting the given values:

$12.08^{2} = 9.72^{2} + 13.91^{2} - 2\left(9.72 \right )\left(13.91 \right )\cos C$

Isolating the angle:

$\cos C = \frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}$

The final step is to take the inverse cosine of both sides:

$C = \cos^{-1}\left(\frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}\right) = 58.31^{\circ} = 58^{\circ}$

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

Find the value of to the nearest tenth.

Answer

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

where , , and are the three sides of the triangle, and is the angle opposite side . Looking at our triangle, taking , then we have , , and . Plugging this into our formula, we get.

Using our calculator to approximate the cosine value gives

Simplifying further gives

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

Solving by taking the square root gives

$x\approx8.6$

Compare your answer with the correct one above

Question

If $\small a=5$ , $\small b=12$ , and $\small c=13$ find $\small \angle C$ to the nearest degree.

Answer

The problem gives the lengths of three sides and asks to find an angle. We can use the Law of Cosines to solve for the angle. Because we are solving for $\small \angle C$ , we use the equation:

$\small \small \small c^2 = a^2 + b^2 - 2abcos\angle C$

Substituting the values from the problem gives

$\small 13^2 = 5^2 + 12^2 - 2(5)(12)cos\angle C$

Isolating $\small \angle C$ by itself gives

$\small \small \angle C = cos(\frac{13^2 - 5^2 - 12^2}{(-2)(5)(12)}) = 90^o$

Compare your answer with the correct one above

Question

If $\small a=4$ , $\small b=3$ , $\small c=5$ , find $\small \angle B$ to the nearest degree.

Answer

We are given the lengths of the three sides to a triangle. Therefore, we can use the Law of Cosines to find the angle being asked for. Since we are looking for $\small \angle B$ we use the equation,

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

Inputting the values we are given,

$\small 3^2 = 4^2 + 5^2 - 2(4)(5)\cos\angle B$

Next we isolate $\small \angle B$ by itself to solve for it

$\small \angle B = \cos^{-1}(\frac{3^2-4^2-5^2}{(-2)(4)(5)})$

$\small \angle B = 37^o$

Compare your answer with the correct one above

Question

If $\small a=10$ , $\small b=6.2$ , and $\small c=4.8$ , find $\small \angle A$ to the nearest degree.

Answer

Because the problem provides all three sides of the triangle, we can use the Law of Cosines to solve this problem. Since we are solving for $\small \angle A$ , we use the equation

$\small a^2 = b^2 + c^2 - 2ac\cos\angle A$

Substitute in the given values

$\small 10^2 = 6.2^2 + 4.8^2 - 2(6.2)(4.8)\cos\angle A$

Isolate $\small \angle A$

$\small \angle A =\cos^{-1}( \frac{10^2-6.2^2-4.8^2}{(-2)(6.2)(4.8)})$

$\small \angle A = 130^o$

Compare your answer with the correct one above

Question

If $\small a=10$ , $\small b=6.7$ , $\small \angle C$ = $\small 45^o$ find $\small c$ to the nearest tenth.

Answer

Because we are given two sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of side $\small c$ . To find side $\small c$ we use

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

Taking the square root of both sides gives us

$\small c = \sqrt{a^2+b^2-2ab\cos\angle C}$

Substituting in the values from the problem

$\small c = \sqrt{10^2 +6.7^2 - 2(10)(6.7)\cos45^o}$

$\small c = 7.1$

Compare your answer with the correct one above

Question

If $\small a=8.4$ , $\small b=11.2$ , $\small \angle C$ = $\small 22^o$ , find the length of side $\small c$ to the nearest tenth of a degree.

Answer

Since we are give the length of two sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of the third side. Because we are looking for , we use the equation

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

Taking the square root of both sides we isolate

$\small c = \sqrt{a^2+b^2-2ab\cos\angle C}$

Substituting in the values from the problem gives

$\small c = \sqrt{8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos\angle 22^o}$

$\small \small c = 4.6$

Compare your answer with the correct one above

Question

If $\small \angle A$ = $\small 20^o$ , $\small b$ = $\small 9$ , $\small c$ = $\small 20$ , find the length of side $\small a$ to the nearest whole number.

Answer

Because the problem gives the length of two of the three sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of the third side. This gives use the equation

$\small a^2 = b^2 + c^2 - 2bc\cos\angle A$

We can take the square root of both sides in order to give the equation

$\small a= \sqrt{b^2+c^2-2bc\cos\angle A}$

Substituting in the values given from the problem gives

$\small a = \sqrt{9^2+20^2-2(9)(20)\cos20^o}$

$\small a = 12$

Compare your answer with the correct one above

Question

Suppose there was a triangle with side lengths 8,7, 14. What is the measure of the largest angle, rounded to the nearest degree?

Answer

The largest angle will be opposite the largest of the three side lengths, 14. We can find its measure using the law of cosines:

$14^2 = 7^2 + 8^2 - 2(7)(8)(cos\ C)$

$196 = 49 + 64 - 112cos \ C = 113 - 112 cos \ C$

$83 = -112 cos \ C$

$-0.74107 = cos \ C$

From here take the inverse cosine of each side to find the degress value that is angle C.

Compare your answer with the correct one above

Question

Solve for x:

Answer

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:

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

Compare your answer with the correct one above

Question

Solve for $\theta$ :

Answer

Solve using law of cosines: $c^2 = a^2 + b^2 - 2ab \cdot \cos C$ where C is the angle between sides a and b.

$8^2 = 3^2 + 10^2 - 2(3)(10) \cos \theta$

$64 = 109 - 60 \cos \theta$ subtract 109 from both sides

$- 45 = -60 \cos \theta$ divide by -60

$0.75 = \cos \theta$ take the inverse cosine using a calculator

$41.41 = \theta$

Sometimes with law of cosines we have to worry about a second angle that the calculator won't give us. In this case, that would be but that is too big an angle to be in a triangle.

Compare your answer with the correct one above

Question

Find the missing angles and sides.

Answer

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°.

Compare your answer with the correct one above

Question

In the triangle below, , , and . Find the measure of $\angle A$ to the nearest tenth.

Answer

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

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

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

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

$A\approx86.6^\circ$

Compare your answer with the correct one above

Question

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

Answer

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

So:

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

$b\approx9.0$

Compare your answer with the correct one above

Question

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?

Answer

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

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

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

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

$c\approx696$

Compare your answer with the correct one above

Tap the card to reveal the answer