Triangles - High School Math

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Question

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?

Answer

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:

, ,

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

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

$51= -408 \cos C$

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

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

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Question

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

Answer

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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Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

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

Substitute the sidelengths and solve for $\cos m \angle A$ :

$32^{2} = 26^{2} + 23^{2} - 2\cdot 26 \cdot 23 \cdot \cos m \angle A$

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx \cos^{-1} 0.1513 \approx 81.3 ^{\circ }$

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Question

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

Answer

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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Question

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

Answer

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:

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

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

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

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

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

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Question

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

Answer

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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Question

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

(NOTE: Figure not necessarily drawn to scale.)

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Answer

We are looking for $\sin(\theta )$ . Remember the definition of in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

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Question

In this figure, if angle $c=90^\circ$ , side , and side , what is the measure of angle ?

Answer

Since $c=90^\circ$ , we know we are working with a right triangle.

That means that $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ .

In this problem, that would be:

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

Plug in our given values:

$\sin(a)=\frac{8}{12}$

$\sin(a)=\frac{2}{3}$

$a=\sin^{-1}(\frac{2}{3})$

$a=41.8^\circ$

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Question

In this figure, side , , and $c=90^\circ$ . What is the value of angle ?

Answer

Since $c=90^\circ$ , we know we are working with a right triangle.

That means that $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$ .

In this problem, that would be:

$\cos(a)=\frac{\text{Y}}{\text{X}}$

Plug in our given values:

$\cos(a)=\frac{\text{13}}{\text{20}}$

$\cos(a)=0.65$

$a=\cos^{-1}(0.65)$

$a=49.46^\circ$

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Question

What is the length of CB?

Answer

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

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

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

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Question

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

Answer

This problem can be easily solved using trig identities. We are given the hypotenuse 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 then you utilized the $\cos$ function rather than the $\sin$ .

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Question

Solve for .

(Figure not drawn to scale).

Answer

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 . In the smaller triangle, the bottom edge has a length of , and in the larger triangle is has a length of . 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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Question

Solve for .

(Figure not drawn to scale).

Answer

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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Question

In this figure, , $Y=6\sqrt{3}$ , and . What is the value of angle ?

Answer

Notice that these sides fit the pattern of a 30:60:90 right triangle: $x:x\sqrt{3}:2x$ .

In this case, .

Since angle is opposite , it must be $30^\circ$ .

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Question

A triangle has angles of $30^\circ:60^\circ:90^\circ$ . If the side opposite the $30^\circ$ angle is , what is the length of the side opposite $60^\circ$ ?

Answer

The pattern for $30^\circ:60^\circ:90^\circ$ is that the sides will be $x:x\sqrt{3}:2x$ .

If the side opposite $30^\circ$ is , then the side opposite $60^\circ$ will be $7\sqrt{3}$ .

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