Triangles - Math

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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 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}$$

Rearrange to solve for .

$\dpi{100} \rightarrow 12.3ft$

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

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 :

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

, ,

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

$51= -408 \cos C$

$\cos C = -$\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 $\cos m \angle A$ :

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

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

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

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

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

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Question

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 and is , how long is ?

Tap to reveal answer

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}$$

Rearrange to solve for .

$\dpi{100} \rightarrow 12.3ft$

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

← Didn't Know|Knew It →

Question

Solve for .

(Figure not drawn to scale).

Tap to reveal answer

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

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}$$

$b=56.44^\circ$

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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}$$

$b=10.41^\circ$

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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 and that have any given positive sine other than 1 - we get the other by subtracting the previous result from :

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

This, however, is impossible, since this would result in the sum of the triangle measures being greater than . This leaves as the only possible answer.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

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Answer

The altitude perpendicular to the hypotenuse of a right triangle divides that triangle into two smaller triangles similar to each other and the large triangle. Therefore, the sides are in proportion. The hypotenuse of the triangle is equal to

Therefore, we can set up, and solve for in, a proportion statement involving the shorter side and hypotenuse of the large triangle and the larger of the two smaller triangles:

$\frac{N}{12}$ = $\frac{5}{13}$

$$\frac{N}{12}$ \times 12 = $\frac{5}{13}$ \times 12$

$N = $\frac{60}{13}$ = 4$\frac{8}{13}$$

This is not one of the choices.

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Question

What is the length of the diagonals of trapezoid ? Assume the figure is an isoceles trapezoid.

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Answer

To find the length of the diagonal, we need to use the Pythagorean Theorem. Therefore, we need to sketch the following triangle within trapezoid :

We know that the base of the triangle has length . By subtracting the top of the trapezoid from the bottom of the trapezoid, we get:

Dividing by two, we have the length of each additional side on the bottom of the trapezoid:

$$\frac{6: m}{2}$ = 3: m$

Adding these two values together, we get .

The formula for the length of diagonal uses the Pythagoreon Theorem:

, where is the point between and representing the base of the triangle.

Plugging in our values, we get:

$AC = $\sqrt{97}$: m$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate the length of the hypotenuse of the blue triangle.

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Answer

The inscribed rectangle is a 20 by 20 square. Since opposite sides of the square are parallel, the corresponding angles of the two smaller right triangles are congruent; therefore, the two triangles are similar and, by definition, their sides are in proportion.

The small top triangle has legs 10 and 20. Therefore, the length of its hypotenuse can be determined using the Pythagorean Theorem:

$c = $$\sqrt{10^{2}$$+ $20^{2}$} = $\sqrt{100 + 400}$ = $\sqrt{500}$ = $\sqrt{100}$ \cdot $\sqrt{5}$ = 10 $\sqrt{5}$$

The small top triangle has short leg 10 and hypotenuse $10 $\sqrt{5}$$ . The blue triangle has short leg 20 and unknown hypotenuse , where can be calculated with the proportion statement

$\frac{N}{10\sqrt{5}$} = $\frac{20}{10}$

$\frac{N}{10\sqrt{5}$} \cdot 10$\sqrt{5}$ = $\frac{20}{10}$ \cdot 10$\sqrt{5}$

$N = 20 $\sqrt{5}$$

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