Trigonometry - SAT Subject Test in Math II

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Question

In $\bigtriangleup ABC$ :

Evaluate $m \angle A$ to the nearest degree.

Answer

The figure referenced is below:

By the Law of Cosines, the relationship of the measure of an angle $\gamma$ of a triangle and the three side lengths , , and , the sidelength opposite the aforementioned angle, is as follows:

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

All three side lengths are known, so we are solving for $\gamma$ . Setting

, the length of the side opposite the unknown angle;

;

;

and $\gamma = \cos A$ ,

We get the equation

$25^{2} =23^{2} +34^{2} - 2 (23) (34)\cos A$

$625 =529+1,156- 1,564 \cos A$

$625 =1,685 - 1,564 \cos A$

Solving for $\cos A$ :

$625 - 1,685 =1,685 - 1,564 \cos A - 1,685$

$-1.060 = - 1,564 \cos A$

$\frac{-1.060 }{- 1,564}= \frac{- 1,564 \cos A}{- 1,564}$

$\cos A \approx 0.6777$

Taking the inverse cosine:

$A = \cos ^{-1} 0.6777 \approx 47 ^{\circ }$ ,

the correct response.

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Question

Give the length of one side of a regular pentagon whose diagonals measure 10 each. (Nearest tenth)

Answer

Construct the pentagon with diagonal $\overline{AC}$ .

We will concern ourselves with finding the length of $\overline{BC}$ .

Since $m \angle ABC = \frac{(5-2)\cdot 180^{ \circ}}{5}= 108 ^{ \circ}$ , and $\Delta ABC$ is isosceles, then

$m \angle BAC = \frac{1}{2} (180^{\circ } - 108^{\circ }) = 36^{\circ }$

The following diagram is formed:

By the Law of Sines,

$\frac{BC}{\sin \left (m \angle BAC \right )} = \frac{AC} {\sin \left (m \angle ABC \right )}$

$\frac{BC}{\sin 36^{\circ }} = \frac{10} {\sin 108^{\circ }}$

$BC= \frac{10\sin 36^{\circ }} {\sin 108^{\circ }}$

$BC\approx \frac{10\cdot 0.5878} {0.9511}$

$BC \approx 6.2$

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Question

A decagon is a ten-sided polygon.

Decagon has diagonal $\overline{AC}$ with length 10. To the nearest tenth, give the length of one side.

Answer

Construct the decagon with diagonal $\overline{AC}$ .

We will concern ourselves with finding the length of $\overline{BC}$ .

Since $m \angle ABC = \frac{(10-2)\cdot 180^{ \circ}}{10}= 144 ^{ \circ}$ , and $\Delta ABC$ is isosceles, then

$m \angle BAC = \frac{1}{2} (180^{\circ } - 144^{\circ }) = 18^{\circ }$

The following diagram is formed (limiting ourselves to $\Delta ABC$ ):

By the Law of Sines,

$\frac{BC}{\sin \left (m \angle BAC \right )} = \frac{AC} {\sin \left (m \angle ABC \right )}$

$\frac{BC}{\sin 18^{\circ }} = \frac{10} {\sin 144^{\circ }}$

$BC= \frac{10\sin 18^{\circ }} {\sin 144^{\circ }}$

$BC\approx \frac{10\cdot 0.3090} {0.5878}$

$BC \approx 5.3$

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Question

A nonagon is a nine-sided polygon.

Nonagon has diagonal $\overline{AC}$ with length 10. To the nearest tenth, give the length of one side.

Answer

Construct the nonagon with diagonal $\overline{AC}$ .

We will concern ourselves with finding the length of $\overline{BC}$ .

Since $m \angle ABC = \frac{(9-2)\cdot 180^{ \circ}}{9}= 140 ^{ \circ}$ , and $\Delta ABC$ is isosceles, then

$m \angle BAC = \frac{1}{2} (180^{\circ } - 140^{\circ }) = 20^{\circ }$

The following diagram is formed (limiiting ourselves to $\Delta ABC$ ):

By the Law of Sines,

$\frac{BC}{\sin \left (m \angle BAC \right )} = \frac{AC} {\sin \left (m \angle ABC \right )}$

$\frac{BC}{\sin 20^{\circ }} = \frac{10} {\sin 140^{\circ }}$

$BC= \frac{10\sin 20^{\circ }} {\sin 140^{\circ }}$

$BC\approx \frac{10\cdot 0.3420} {0.6428}$

$BC \approx 5.3$

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Question

The area of a regular pentagon is 1,000. Give its perimeter to the nearest whole number.

Answer

A regular pentagon can be divided into ten congruent triangles by its five radii and its five apothems. Each triangle has the following shape:

The area of one such triangle is $\frac{1}{2}ab$ , so the area of the entire pentagon is ten times this, or $10 \cdot \frac{1}{2}ab = 5ab$ .

The area of the pentagon is 1,000, so

Also,

$\frac{a}{b} = \tan 54^{\circ }$ , or equivalently, $a=b \tan 54^{\circ }$ , so we solve for in the equation:

$b \tan 54^{ \circ } \cdot b = 200$

$b^{2} \tan 54^{ \circ } = 200$

$b^{2} = \frac{200}{\tan 54^{ \circ }} \approx \frac{200}{ 1.3764} \approx 145.3$

$b \approx \sqrt{145.3} \approx 12.1$

The perimeter is ten times this, or 121.

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Question

The area of a regular dodecagon (twelve-sided polygon) is 600. Give its perimeter to the nearest whole number.

Answer

A regular dodecagon can be divided into twenty-four congruent triangles by its twelve radii and its twelve apothems, each of which is shaped as shown:

The area of one such triangle is $\frac{1}{2}ab$ , so the area of the entire dodecagon is twenty-four times this, or

$24 \cdot \frac{1}{2}ab = 12ab$ .

The area of the dodecagon is 600, so

, or

.

Also,

$\frac{a}{b} = \tan 75^{\circ }$ , or equivalently, $a=b \tan 75^{\circ }$ , so solve for in the equation

$b \tan 75^{ \circ } \cdot b =50$

Solve for :

$b^{2} \cdot \tan 75^{ \circ } = 50$

$b^{2} = \frac{50}{ \tan 75^{ \circ }} \approx \frac{50}{ 3.7321} \approx 13.4$

$b \approx \sqrt{13.4} \approx 3.66$

The perimeter is twenty-four times this:

$P \approx 3.66 \cdot 24 \approx 88$

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Question

The area of a regular nonagon (nine-sided polygon) is 900. Give its perimeter to the nearest whole number.

Answer

A regular nonagon can be divided into eighteen congruent triangles by its nine radii and its nine apothems, each of which is shaped as shown:

The area of one such triangle is $\frac{1}{2}ab$ , so the area of the entire nonagon is eighteen times this, or $18 \cdot \frac{1}{2}ab = 9ab$ . Since the area is 900,

, or

.

Also,

$\frac{a}{b} = \tan 70^{\circ }$ , or equivalently, $a=b \tan 70^{\circ }$ , so solve for in the equation

$b \tan 70^{\circ } \cdot b= 100$

$b ^{2} \cdot \tan 70^{\circ } = 100$

$b ^{2} = \frac{100}{\tan 70^{\circ } } \approx \frac{100}{ 2.7475 } \approx 36.40$

$b \approx \sqrt{36.40} \approx 6.03$

The perimeter of the nonagon is eighteen times this:

$18 \cdot 6.03 \approx 109$ , the correct response.

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Question

In $\bigtriangleup ABC$ :

$m \angle A = 116 ^{\circ }$

$m \angle B = 20^{\circ }$

Evaluate to the nearest whole unit.

Answer

The Law of Sines states that given two angles of a triangle with measures $\alpha, \beta$ , and their opposite sides of lengths , respectively,

$\frac{\sin \alpha}{a} = \frac{\sin \beta}{b}$ ,

or, equivalently,

$\frac{b}{\sin \beta} = \frac{a}{\sin \alpha}$ .

$\overline{AC }$ , whose length is desired, and $\overline{BC}$ , whose length is given, are opposite $\angle B$ and $\angle A$ , respectively, so, in the sine formula, set , $\beta = m \angle B = 20 ^{\circ }$ , , and $\alpha = m \angle A = 116^{\circ }$ in the Law of Sines formula, then solve for :

$\frac{AC}{\sin 20^{\circ }} = \frac{40}{\sin 116^{\circ }}$

$\frac{AC}{\sin 20^{\circ }} \cdot \sin 20^{\circ } = \frac{40}{\sin 116^{\circ }} \cdot \sin 20^{\circ }$

$AC = \frac{40}{\sin 116^{\circ }} \cdot \sin 20^{\circ }$

$\approx \frac{40}{0.8988} \cdot 0.3420$

$\approx 15$

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Question

Suppose the distance from a student's eyes to the floor is 4 feet. He stares up at the top of a tree that is 20 feet away, creating a 30 degree angle of elevation. How tall is the tree?

Answer

The height of the tree requires using trigonometry to solve. The distance of the student to the tree , partial height of the tree , and the distance between the student's eyes to the top of the tree will form the right triangle.

The tangent operation will be best used for this scenario, since we have the known distance of the student to the tree, and the partial height of the tree.

Set up an equation to solve for the partial height of the tree.

$tan(\theta) = \frac{\textup{Opposite side to angle}}{\textup{Adjacent side to angle}}$

$tan(30) = \frac{P}{20}$

Multiply by 20 on both sides.

$P=20tan(30) = 20(\frac{\sqrt{3}}{3}) = \frac{20\sqrt{3}}{3}$

We will need to add this with the height of the student's eyes to the ground to get the height of the tree.

$\frac{20\sqrt{3}}{3} + 4 = \frac{20\sqrt{3}}{3} + \frac{12}{3}$

The answer is: $\frac{20\sqrt{3}+12}{3}$

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Question

A triangle has sides that measure 10, 12, and 16. What is the greatest measure of any of its angles (nearest tenth of a degree)?

Answer

We are seeking the measure of the angle opposite the side of greatest length, 16.

We can use the Law of Cosines, setting , and solving for :

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

$\cos C = \frac{10^{2}+12^{2}- 16^{2}}{2 \cdot 10 \cdot 12}$

$\cos C = \frac{100+144- 256}{240}$

$\cos C = \frac{-12}{240}$

$\cos C =- 0.0500$

$C = \cos ^{-1}(-0.0500) \approx 92.9^{\circ}$

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Question

A triangle has sides that measure 15, 17, and 30. What is the least measure of any of its angles (nearest tenth of a degree)?

Answer

We are seeking the measure of the angle opposite the side of least length, 15.

We can use the Law of Cosines, setting , and solving for :

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

$\cos C = \frac{17^{2}+30^{2}- 15^{2}}{2 \cdot 17 \cdot 30}$

$\cos C = \frac{289+900- 225 }{1,020}$

$\cos C = \frac{964 }{1,020}$

$\cos C \approx 0.9451$

$C \approx \cos ^{-1} 0.9451 \approx 19.1^{\circ}$

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Question

Given : $\Delta ABC$ with $AB = 12, BC = 17, m \angle B= 136 ^{\circ}$ .

Which of the following whole numbers is closest to ?

Answer

Apply the Law of Cosines

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

setting $a = 12, b= 17, C= 136 ^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+17^{2}- 2\cdot 12 \cdot 17 \cos 136^{\circ}}$

$c \approx \sqrt{144+289-408 (-0.7193)}$

$c \approx \sqrt{144+289+293.49}$

$c \approx \sqrt{726.49}$

$c \approx 27.0$

Of the five choices, 27 comes closest.

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Question

Given : $\Delta ABC$ with $AB = 12, BC = 16, m \angle B= 56 ^{\circ}$ .

Evaluate to the nearest tenth.

Answer

Apply the Law of Cosines

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

setting $a = 12, b = 16, C = 56^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+16^{2}- 2\cdot 12 \cdot 16 \cos 56^{\circ}}$

$c \approx \sqrt{ 144+256- 384 \cdot 0.5592 }$

$c \approx \sqrt{ 144+256- 214.73 }$

$c \approx \sqrt{ 185.27 }$

$c \approx 13.6$

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Question

In $\bigtriangleup ABC$ :

$m \angle A = 105^{\circ }$

Evaluate the length of $\overline{BC}$ to the nearest tenth of a unit.

Answer

The figure referenced is below:

By the Law of Cosines, given the lengths and of two sides of a triangle, and the measure $\gamma$ of their included angle, the length of the third side can be calculated using the formula

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

Substituting , , , and $\gamma = m \angle A = 105^{\circ }$ , then evaluating:

$(BC)^{2} = 12^{2} + 15 ^{2} - 2 (12)(15)\cos 105^{\circ }$

$\approx 144+225 - 360 (-0.2588 )$

$\approx 369 - (-93.17 )$

$\approx 369 - (-93.17 )$

$\approx 462.17$

Taking the square root of both sides:

$BC\approx \sqrt{462.17} \approx 21.6$

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Question

In $\bigtriangleup ABC$ ,

$\angle C =65 ^{\circ }$

Evaluate $m \angle A$ (nearest degree)

Answer

By the Law of Sines, if and are the lengths of two sides of a triangle, and $\alpha$ and $\beta$ the measures of their respective opposite angles,

$\frac{\sin \alpha }{a} = \frac{\sin \beta }{b}$

$\angle A$ and $\angle C$ are opposite sides $\overline{BC}$ and $\overline{AB}$ , so, setting $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , , and :

$\frac{\sin \alpha }{34} = \frac{\sin 65^{\circ } }{23}$

$\frac{\sin \alpha }{34}\cdot 34 = \frac{\sin 65^{\circ } }{23} \cdot 34$

$\sin \alpha \approx \frac{0.9063 }{23} \cdot 34 \approx 1.3398$

However, the range of the sine function is , so there is no value of $\alpha$ for which this is true. Therefore, this triangle cannot exist.

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Question

Find the measure of angle .

Answer

Start by using the Law of Sines to find the measure of angle .

$\frac{8}{\sin 39}=\frac{12}{\sin B}$

$12\sin 39=8\sin B$

$\sin B=0.94398$

$B=70.73^{\circ}$

Since the angles of a triangle must add up to ,

$A=70.27^{\circ}$

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Question

The degree angle can be expressed as what in radians?

Answer

In order to convert degrees to radians, we will need to know the conversion factor.

$180 \textup{ deg} = \pi \textup{ rad}$

Set up a dimensional analysis to solve.

$300 \textup{ deg} (\frac{\pi \textup{ rad}}{180 \textup{ deg}}) = \frac{5}{3}\pi\textup{ rad}$

The answer is: $\frac{5}{3}\pi$

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Question

Which of the following angles belong in the fourth quadrant?

Answer

The fourth quadrant is in the positive x-axis and negative y-axis.

The angle ranges are:

$270\rightleftharpoons 360 \textup{ deg}$

$\frac{3}{2}\pi\rightleftharpoons 2\pi \textup{ radians}$

The only possible answer is: $\frac{17}{9}\pi$

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Question

What degree measure is equivalent to $4\pi$ ?

Answer

Every pi radians is equal to 180 degrees.

Replace the pi term with 180 degrees and multiply.

$4\pi = 4(180 \textup{ degrees}) = 720$

The answer is:

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Question

If , what is if is between and $2\pi$ ?

Answer

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .

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