Finding Sides with Trigonometry - SAT Subject Test in Math II

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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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