Triangles - Math

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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 ?

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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 ?

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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 ?

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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

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

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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 $60^\circ$ ?

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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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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?

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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 ?

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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 $m \angle A$ ?

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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 ?

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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 ?

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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 ?

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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.)

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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 ?

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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 $m \angle C$ ?

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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

What is the length of CB?

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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 ?

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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).

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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).

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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

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 ?

Tap to reveal answer

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 ?

Tap to reveal answer

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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