SAT Math - Finding Sides with Trigonometry

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?

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

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

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

$10\sqrt3+4$

Explanation

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: $\frac{20\sqrt{3}+12}{3}$

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

Explanation

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:

Thingy_1

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.

In $\bigtriangleup ABC$ :

$m \angle A = 76 ^{\circ }$

$m \angle B = 65 ^{\circ }$

$\textup{Evaluate}$ $\textup{to the nearest whole unit.}$

Explanation

The figure referenced is below:

Triangle z

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

$\frac{\sin \beta}{b} = \frac{\sin \gamma}{c}$ ,

or, equivalently,

$\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$ .

In this formula, we set:

, the desired sidelength;

$\beta =m \angle B = 65^{\circ }$ , the measure of its opposite angle;

, the known sidelength;

$\gamma =m \angle C$ , the measure of its opposite angle, which is

$m \angle C = 180 ^{\circ } - (m \angle A + m \angle B)$

$= 180 ^{\circ } - ( 76^{\circ } + 65 ^{\circ })$

$= 180 ^{\circ } - 141 ^{\circ }$

$=39 ^{\circ }$

Substituting in the Law of Sines formula and solving for :

$\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$

$\frac{b}{\sin 65 ^{\circ }} = \frac{ 34}{\sin 39 ^{\circ }}$

$\frac{b}{\sin 65 ^{\circ }} \cdot \sin 65 ^{\circ } = \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }$

$b= \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }$

Evaluating the sines, then calculating:

$b\approx \frac{ 34}{0.6293 } \cdot 0.9063$

$b \approx 49$

A plane flies degrees north of east for miles. It then turns and flies degrees south of east for miles. Approximately how many miles is the plane from its starting point? (Ignore the curvature of the Earth.)

Explanation

The plane flies two sides of a triangle. The angle formed between the two sides is 40 degrees. In a Side-Angle-Side situation, it is appropriate to employ the use of the Law of Cosines.

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

$c^2 = 100^2 + 200^2 - 2(100)(200) \cos 40^{\circ}$

$c^2 \approx 19358$

$c \approx \sqrt{19358} \approx 139$

In $\bigtriangleup ABC$ :

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

$112 ^{\circ }$

$68^{\circ }$

$151^{\circ }$

$141^{\circ }$

Insufficient information is provided to answer the question.

Explanation

The figure referenced is below:

Triangle z

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 sidelengths 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

$25^{2} =13^{2} +17^{2} - 2 (13) (17)\cos A$

$625 =169 +289- 442 \cos A$

$625 =458- 442 \cos A$

Solving for $\cos A$ :

$625 - 458 =458- 442 \cos A - 458$

$167 = - 442 \cos A$

$\frac{167}{-442} = \frac{- 442 \cos A}{-442}$

$\cos A \approx -0.3778$

Taking the inverse cosine:

$A = \cos ^{-1} -0.3778 \approx 112 ^{\circ }$ ,

the correct response.

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.

Explanation

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 $\Delta ABC$ ):

Decagon

By the Law of Sines,

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

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

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

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

$BC \approx 5.3$

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

Explanation

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:

Thingy_2

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:

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

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

Explanation

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

Thingy_1

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.

In $\bigtriangleup ABC$ :

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

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

Evaluate to the nearest whole unit.

Explanation

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 :

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

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

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

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

$\approx 15$

In $\bigtriangleup ABC$ :

$m \angle A = 57^{\circ }$

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

Explanation

The figure referenced is below:

Triangle z

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