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Trigonometry : Triangles

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Example Questions

Example Question #22 : Law Of Cosines

A triangle has side lengths a = 1 , b = 1.2 and $\angle C=150^{\circ}$

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

Possible Answers:

$c=\frac{1.2\sin 150^{\circ}}{\sin 30^{\circ}}$

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

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

$c=\frac{1.2\sin 30^{\circ}}{ \sin 150^{\circ}}$

$c=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 30^{\circ}}$

Correct answer:

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

Explanation:

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.

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Example Question #23 : Law Of Cosines

Given a = 6.2 , b = 5.1 and c = 3.5 determine to the nearest degree the measure of $\angle B$ .

Possible Answers:

$55^{\circ}$

$35^{\circ}$

$34^{\circ}$

$56^{\circ}$

$90^{\circ}$

Correct answer:

$55^{\circ}$

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:

$B = 55^{\circ}$

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Example Question #24 : Law Of Cosines

If a = 9.72 , b = 13.91 and c = 12.08 , determine the measure of $\angle C$ to the nearest degree.

Possible Answers:

$58^{\circ}$

$31^{\circ}$

$103^{\circ}$

$59^{\circ}$

$93^{\circ}$

Correct answer:

$58^{\circ}$

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:

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

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Example Question #25 : Law Of Cosines And Law Of Sines

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

Possible Answers:

$\small 110^o$

$\small 73^o$

$\small 37^o$

$\small 90^{o}$

Correct answer:

$\small 90^{o}$

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

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

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Example Question #24 : Law Of Cosines

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

Possible Answers:

$\small 37^o$

$\small 100^o$

$\small 53^o$

$\small 90^o$

Correct answer:

$\small 37^o$

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

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

$\small \angle B = 37^o$

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Example Question #61 : Triangles

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

Possible Answers:

$\small 90^o$

$\small 70^o$

$\small 130^o$

$\small 105^o$

Correct answer:

$\small 130^o$

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 $\small \angle A$

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

$\small \angle A = 130^o$

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Example Question #62 : Triangles

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

Possible Answers:

$\small 4.0$

$\small 8.7$

$\small 7.1$

$\small 2.3$

Correct answer:

$\small 7.1$

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

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

$\small c = 7.1$

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Example Question #63 : Triangles

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.

Possible Answers:

$\small 9.8$

$\small 5.3$

$\small 7.2$

$\small 4.6$

Correct answer:

$\small 4.6$

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

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

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

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

$\small \small c = 4.6$

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Example Question #31 : Law Of Cosines

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.

Possible Answers:

$\small 12$

$\small 10$

$\small 7$

$\small 5$

Correct answer:

$\small 12$

Explanation:

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

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

$\small a = 12$

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Example Question #33 : Law Of Cosines And Law Of Sines

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

Possible Answers:

123^o

98^o

52^o

158^o

138^o

Correct answer:

138^o

Explanation:

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.

C = 138^o

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Trigonometry : Triangles

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #22 : Law Of Cosines

Example Question #23 : Law Of Cosines

Example Question #24 : Law Of Cosines

Example Question #25 : Law Of Cosines And Law Of Sines

Example Question #24 : Law Of Cosines

Example Question #61 : Triangles

Example Question #62 : Triangles

Example Question #63 : Triangles

Example Question #31 : Law Of Cosines

Example Question #33 : Law Of Cosines And Law Of Sines

All Trigonometry Resources

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