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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Applying The Law Of Cosines

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

Possible Answers:

$86.4^{\circ }$

$119.2^{\circ }$

$97.2^{\circ }$

$93.6^{\circ }$

$130.1^{\circ }$

Correct answer:

$97.2^{\circ }$

Explanation:

We can apply the Law of Cosines to find the measure of this angle, which we will call :

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

The widest angle will be opposite the side of length 22, so we will set:

c=22 , a=12 , b=17

$22^{2} = 12^{2} + 17^{2} -2\cdot 12\cdot 17 \cos C$

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -\frac{51}{408} = -0.125$

$C = \cos^{-1} 0.125 = 97.2^{\circ }$

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Example Question #1821 : High School Math

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , AB = 26 , and AC = 23 . To the nearest tenth, what is ?

Possible Answers:

112.4

26.8

41.1

A triangle with these characteristics cannot exist.

47.9

Correct answer:

26.8

Explanation:

By the Law of Cosines:

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

or, equivalently,

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

Substitute:

$BC =\sqrt{ 26^{2} +23^{2} - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx \sqrt{ 676 +529 - 1,196 \cdot 0.4067}$

$\approx \sqrt{ 718.6} \approx 26.8$

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Example Question #1822 : High School Math

Rt_triangle_letters
In this figure, angle $a=30^\circ$ and side Z=15 . If angle $b=45^\circ$ , what is the length of side ?

Possible Answers:

$\frac{15\sqrt{2}}{2}$

$30\sqrt{2}$

$15\sqrt{2}$

7.5

Correct answer:

$15\sqrt{2}$

Explanation:

For this problem, use the law of sines:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$ .

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{15}{\sin(30^\circ)}=\frac{Y}{\sin{(45^\circ)}}$

$\frac{15}{\frac{1}{2}}=\frac{Y}{\frac{\sqrt{2}}{2}}$

Cross multiply:

$15*\frac{\sqrt{2}}{2}=\frac{1}{2}Y$

Multiply both sides by :

$15\sqrt{2}=Y$

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Example Question #1823 : High School Math

Rt_triangle_letters

In this figure $a=22^\circ$ and $c=85^\circ$ . If X=30 , what is ?

Possible Answers:

20.1

11.28

30.09

0.374

0.997

Correct answer:

11.28

Explanation:

For this problem, use the law of sines:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$ .

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }c}{\sin(c)}$

$\frac{Z}{\sin(a)}=\frac{X}{\sin{(c)}}$

$\frac{Z}{\sin(22^\circ)}=\frac{30}{\sin{85^\circ}}$

$\frac{Z}{0.375}=\frac{30}{0.997}$

$\frac{Z}{0.375}=30.09$

Z=30.09*0.375

Z=11.28

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Example Question #95 : Trigonometry

In $\Delta ABC$ , $m \angle A = 75^{\circ }$ , $m \angle B = 62^{\circ }$ , and BC = 16 . To the nearest tenth, what is ?

Possible Answers:

14.6

27.9

11.3

22.7

17.5

Correct answer:

11.3

Explanation:

Since we are given and want to find , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$

$m \angle A = 75^{\circ }$ and

$m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$

Substitute in the above equation:

$\frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$

Cross-multiply and solve for :

$AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$

$AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$

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Example Question #96 : Trigonometry

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , AB = 16 , and BC = 23 . To the nearest tenth, what is $m \angle C$ ?

Possible Answers:

$39.5^{\circ }$

No triangle can exist with these characteristics.

$73.6^{\circ } \textrm{ or }106.4^{\circ }$

$73.6^{\circ }$

$39.5^{\circ } \textrm{ or } 140.5^{\circ }$

Correct answer:

$39.5^{\circ }$

Explanation:

Since we are given $m \angle A$ , , and , and want to find $m \angle C$ , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$ .

Substitute and solve for $\sin m \angle C$ :

$\frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $0^{\circ }$ and $180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $180^{\circ }$ :

$180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $180^{\circ }$ . This leaves $39.5^{\circ }$ as the only possible answer.

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Example Question #3 : Graphs And Inverses Of Trigonometric Functions

Rt_triangle_letters

In this figure, angle $a=30^\circ$ . If side Z =12 and Y=20 , what is the value of angle ?

Possible Answers:

$24^\circ$

$90^\circ$

Undefined

$56.44^\circ$

$0.83^\circ$

Correct answer:

$56.44^\circ$

Explanation:

For this problem, use the law of sines:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$ .

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{12}{\sin(30^\circ)}=\frac{20}{\sin{(b)}}$

$24=\frac{20}{\sin(b)}$

$\sin(b)=\frac{20}{24}$

$b=\sin^{-1}(\frac{20}{24})$

$b=56.44^\circ$

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Example Question #1822 : High School Math

Rt_triangle_letters In this figure, if angle $a=18.5^\circ$ , side Z =30.2 , and side Y=17.2 , what is the value of angle ?

(NOTE: Figure not necessarily drawn to scale.)

Possible Answers:

$71.5^\circ$

Undefined

$33.6^\circ$

$61.22^\circ$

$10.41^\circ$

Correct answer:

$10.41^\circ$

Explanation:

First, observe that this figure is clearly not drawn to scale. Now, we can solve using the law of sines:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}$ .

In this case, we have values that we can plug in:

$\frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}$

$\frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}$

$\frac{30.2}{\sin(18.5^\circ)}=\frac{17.2}{\sin{(b)}}$

$95.18=\frac{17.2}{\sin(b)}$

$\sin(b)=\frac{17.2}{95.18}$

$b=\sin^{-1}(\frac{17.2}{95.18})$

$b=10.41^\circ$

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Example Question #1 : Finding Sides

Solve for .

(Figure not drawn to scale).

Possible Answers:

There is not enough information

2.5

Correct answer:

2.5

Explanation:

The side-angle-side (SAS) postulate can be used to determine that the triangles are similar. Both triangles share the angle farthest to the right. In the smaller triangle, the upper edge has a length of , and in the larger triangle is has a length of 8+8=16 . In the smaller triangle, the bottom edge has a length of , and in the larger triangle is has a length of 6+6=12 . We can test for comparison.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}$

$\frac{8}{16}=\frac{6}{12}=\frac{1}{2}$

The statement is true, so the triangles must be similar.

We can use this ratio to solve for the missing side length.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}=\frac{x}{Left\ edge_2}$

To simplify, we will only use the lower edge and left edge comparison.

$\small \frac{6}{12}=\frac{x}{5}$

Cross multiply.

$\small 12x=30$

$\small x=\frac{30}{12}=\frac{5}{2}=2.5$

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Example Question #1825 : High School Math

Solve for .

(Figure not drawn to scale).

Possible Answers:

16.7

7.3

44.6

6.7

Correct answer:

7.3

Explanation:

We can solve using the trigonometric definition of tangent.

$\small tan\Theta =\frac{opp}{adj}$

We are given the angle and the adjacent side.

$\small tan(22^o)=\frac{z}{18}$

We can find $\small tan(22^o)$ with a calculator.

$\small tan(22^o)=0.4$

$0.4=\frac{z}{18}$

$\small 18*0.4=z$

$\small z=7.3$

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Applying The Law Of Cosines

Example Question #1821 : High School Math

Example Question #1822 : High School Math

Example Question #1823 : High School Math

Example Question #95 : Trigonometry

Example Question #96 : Trigonometry

Example Question #3 : Graphs And Inverses Of Trigonometric Functions

Example Question #1822 : High School Math

Example Question #1 : Finding Sides

Example Question #1825 : High School Math

All High School Math Resources

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