Trigonometry - Law of Cosines

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

$55^{\circ}$

$56^{\circ}$

$34^{\circ}$

$35^{\circ}$

$90^{\circ}$

Explanation

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:

$B = 55^{\circ}$

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

$58^{\circ}$

$59^{\circ}$

$31^{\circ}$

$93^{\circ}$

$103^{\circ}$

Explanation

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:

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

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

Pythagorean Theorem

Vertical Angle Theorem

Isosceles Triangle Theorem

Mean Value Theorem

Cosine 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.

Find the value of to the nearest tenth.

Explanation

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

$x\approx8.6$

Screen_shot_2015-03-07_at_1.33.25_pm

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

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:

Which yields

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

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

$\small 90^{o}$

$\small 37^o$

$\small 110^o$

$\small 73^o$

Explanation

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

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

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

$\small 37^o$

$\small 53^o$

$\small 90^o$

$\small 100^o$

Explanation

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

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

$\small \angle B = 37^o$

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

$\small 130^o$

$\small 90^o$

$\small 70^o$

$\small 105^o$

Explanation

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 $\small \angle A$

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

$\small \angle A = 130^o$

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

$\small 7.1$

$\small 2.3$

$\small 4.0$

$\small 8.7$

Explanation

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

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

$\small c = 7.1$

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.

$\small 4.6$

$\small 7.2$

$\small 9.8$

$\small 5.3$

Explanation

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