Law of Cosines and Sines - Pre-Calculus

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Question

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Answer

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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Question

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Answer

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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Question

Which is NOT an angle of the following triangle?

(Not drawn to scale.)

Answer

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.

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

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

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

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

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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Question

Use the Law of Sines to find in the following triangle.

(Not drawn to scale.)

Answer

We use the Law of Sines to solve this problem

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

we plug in $b=12, B=70^{\circ}, C=45^{\circ}$

$\frac{12}{\sin 70^{\circ}}=\frac{c}{\sin 45^{\circ}}$

solving for we get:

$c=\frac{12\times \sin 45^{\circ}}{\sin 70^{\circ}}=9.03$

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Question

Use the Law of Sines to find .

(Not drawn to scale.)

Answer

We use the Law of Sines to solve this problem:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

where $A=65^{\circ}, B=30^{\circ}, C=85^{\circ}, a=13$

We plug in the values that we will need:

$\frac{13}{\sin 65^{\circ}}=\frac{b}{\sin 30^{\circ}}$

Notice that we did not use .

Solve for we get:

$b=\frac{13\times \sin 30^{\circ}}{\sin 65^{\circ}}=7.17$

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Question

Which of the following are the missing sides of the triangle?

$\text{I. } 13.95$

$\text{II. }6.55$

$\text{III. }4.59$

$\text{IV. }9.77$

Answer

In order to solve this problem, we need to find . We do so by remembering the sum of the angles in a triangle is $180^{\circ}$ :

$C=180^{\circ}-55^{\circ}-35^{\circ}=90^{\circ}$

We can now use the Law of Sines to find the missing sides.

$\frac{a}{\sin 55}=\frac{8}{\sin 90}$

$a=\frac{8\times \sin 55}{\sin 90}=6.55$ which is II.

$\frac{b}{\sin 35}=\frac{8}{\sin 90}$

$b=\frac{8\times \sin 35}{\sin 90}=4.59$ which is III.

Our answers are then II and III

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Question

Which of the following are the missing sides of the triangle?

$\text{I. }4.55$

$\text{II. }5.49$

$\text{III. }1.72$

$\text{IV. }14.56$

Answer

In order to solve this problem, we need to find

Since all the angles of a triangle add to $180^{\circ}$ , we can easily find it:

$180^{\circ}-65^{\circ}-20^{\circ}=95^{\circ}$

We can now use the Law of Sines to find the missing sides:

$\frac{a}{\sin 65}=\frac{5}{\sin 95}$

$a=\frac{5\times \sin 65}{\sin 95}=4.55$ which is I.

$\frac{b}{\sin 20}=\frac{5}{\sin 95}$

$b=\frac{5\times \sin 20}{\sin 95}=1.72$ which is III.

Our answers are then I and III.

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Question

Solve the triangle

Answer

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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Question

Solve the triangle using the Law of Sines:

Answer

First we need to know what the Law of Sines is:

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

Looking at the triangle, we know c, C, and B. We can either solve for side b, using the law, or angle A using our knowledge that the interior angles of a triangle must add up to be 180.

$\frac{b}{sin50^{\circ}} = \frac{12}{sin110^{\circ}}$

$b = \frac{12\cdot sin{50^{\circ}}}{sin110^{\circ}} = 9.78$

$A = 180^{\circ} - 110^{\circ} - 50^{\circ} = 20^{\circ}$

Now all that's left is to find side a:

$\frac{a}{sin20^{\circ}} = \frac{12}{sin110^{\circ}}$

$a = \frac{12\cdot sin{20^{\circ}}}{sin110^{\circ}} = 4.37$

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Question

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

Answer

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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Question

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?

Answer

Write the formula for the Law of Cosines.

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

Round this to the nearest integer.

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Question

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 $30^\circ$ ?

Answer

Write the formula for the Law of Cosines.

Substitute the known values and solve for .

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

$c^2=500-200\sqrt3$

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

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Question

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

Answer

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,

.

Then take the square of both sides: .

Finally, round to the appropriate units: .

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Question

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

Answer

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

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, .

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

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

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Question

Given and $B=36^\circ$ , what is the measurement of to the nearest degree?

Answer

Using the information we have, we can solve for $\angle C$ :

$\frac{sinB}{b}=\frac{sinC}{c}$ .

Plugging in what we know, we have:

$\frac{sin(36)}{4}=\frac{sinC}{6}$ .

Then, solve for :

$6\cdot\frac{sin(36)}{4}=sinC$ .

Simplify, then solve for : which means $sin^{-1}(.8816778784)=C$ .

Therefore, after rounding to the nearest degree, $C=62^\circ$ .

To solve for $\angle A$ , subtract $\angle B$ and $\angle C$ from $180^\circ$ : $180^\circ-36^\circ-62^\circ=82^\circ$ .

Therefore, $\angle A = 82^\circ$ .

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Question

What is the largest possible angle, measured in degrees, in triangle if $\angle A = 30^\circ$ , , and ?

Answer

In the ambiguous SSA case, use Law of Sines to solve for the angle opposite the given side.

$\* \frac {\sin A}{a} = \frac {\sin C}{c} \* \* \frac {\sin 30^\circ}{6} = \frac {\sin C}{10} \* \* C = \arcsin \left(\frac {10 \sin 30^\circ}{6}\right) \approx 56.44^\circ$

If $C = 56.44^\circ$ , then $B = 93.56^\circ$ .

However! Another possible value of C is $(180 - 56.44)^\circ = 123.56^\circ$ .

In this case the angles will be $30^\circ, 123.56^\circ, 26.44^\circ$ .

This is bigger than $93.56^\circ$ and is consequently the answer.

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Question

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

Answer

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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Question

Find the length of the missing side, .

Answer

First, use the Law of Sines to find the measurement of angle

$\frac{6}{\sin34}=\frac{9}{\sin B}$

$6 \sin B = 5.03$

$\sin B=0.84$

Recall that all the angles in a triangle need to add up to degrees.

Now, use the Law of Sines again to find the length of .

$\frac{d}{\sin 88.86}=\frac{6}{\sin 34}$

$d \sin 34=5.9988$

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Question

Find the length of the missing side, .

Answer

First, use the Law of Sines to find the measurement of angle

$\frac{15}{\sin 41}=\frac{20}{\sin B}$

$15\sin B = 13.12$

$\sin B=0.87$

Recall that all the angles in a triangle need to add up to degrees.

Now, use the Law of Sines again to find the length of .

$\frac{d}{\sin 77.99}=\frac{15}{\sin 41}$

$d \sin 41= 14.67$

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Question

Find the length of the missing side, .

Answer

First, use the Law of Sines to find the measurement of angle

$\frac{12}{\sin 21}=\frac{16}{\sin B}$

$12\sin B = 5.73$

$\sin B=0.48$

Recall that all the angles in a triangle need to add up to degrees.

Now, use the Law of Sines again to find the length of .

$\frac{d}{\sin 130.31}=\frac{12}{\sin 21}$

$d \sin 21= 9.15$

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