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High School Math : Triangles

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Example Question #3 : Applying The Law Of Sines

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

Possible Answers:

17.5

27.9

11.3

22.7

14.6

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 #1 : Applying The Law Of Sines

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

Possible Answers:

$73.6^{\circ }$

$39.5^{\circ }$

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

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

No triangle can exist with these characteristics.

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 #61 : Pre Calculus

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:

$90^\circ$

$0.83^\circ$

$56.44^\circ$

$24^\circ$

Undefined

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 #6 : Applying The Law Of Sines

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:

$61.22^\circ$

$71.5^\circ$

$10.41^\circ$

$33.6^\circ$

Undefined

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 #2 : Solving Triangles

Solve for .

(Figure not drawn to scale).

Possible Answers:

44.6

6.7

16.7

7.3

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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Example Question #3 : Solving Triangles

Trig_id

If $\theta$ equals $55^{\circ}$ and is 15ft , how long is ?

Possible Answers:

8.6ft

12.3ft

Not enough information to solve

13.2ft

11.3ft

Correct answer:

12.3ft

Explanation:

This problem can be easily solved using trig identities. We are given the hypotenuse (15ft) and $\theta (55^{\circ})$ . We can then calculate side using the $\sin$ .

$\sin=\frac{opposite}{hypotenuse}$

$\sin55^{\circ}=\frac{o}{15ft}$

Rearrange to solve for .

$o=\sin55^{\circ}*15ft$

$\dpi{100} \rightarrow 12.3ft$

If you calculated the side to equal 8.6ft then you utilized the $\cos$ function rather than the $\sin$ .

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Example Question #4 : Solving Triangles

Triangle

What is the length of CB?

Possible Answers:

$\frac{AC}{\sin A}$

$\frac{AB}{\sin A}$

$\frac{\sin B}{AC}$

$\frac{AB}{\tan C}$

$\frac{\tan C}{AB}$

Correct answer:

$\frac{AB}{\tan C}$

Explanation:

$\tan X= \frac{OPPOSITE}{ADJACENT}$

$\tan C=\frac{AB}{CB}$

$CB= \frac{AB}{\tan C}$

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High School Math : Triangles

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Applying The Law Of Sines

Example Question #1 : Applying The Law Of Sines

Example Question #61 : Pre Calculus

Example Question #6 : Applying The Law Of Sines

Example Question #1 : Finding Sides

Example Question #2 : Solving Triangles

Example Question #3 : Solving Triangles

Example Question #4 : Solving Triangles

All High School Math Resources

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