Trigonometry - SAT Subject Test in Math I

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Question

What is the measure of the angle made between a line segment with points , and the -axis? Round your answer to the nearest hundreth of a degree.

Answer

Based on the information given, we know that the ratio of to on this segment could be represented as:

$\frac{8}{5}$

This ratio represents the tangent of the triangle formed by our line segment and the -axis. Using the inverse tangent function, we can find the angle measure:

$\tan^{-1}(\frac{8}{5}) = 57.9946167919165$

This refers to a reference angle of $58.00^{\circ}$

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Question

What is the measure of the angle made between a line segment with points , and the -axis? Round your answer to the nearest hundreth of a degree.

Answer

Based on the information given, we know that the ratio of to on this segment could be represented as:

$\frac{-8}{7}$

This ratio represents the tangent of the triangle formed by our line segment and the -axis. Using the inverse tangent function, we can find the angle measure:

$\tan^{-1}(\frac{-8}{7}) = -48.81407483429027$

This refers to a reference angle of $48.81^{\circ}$ .

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Question

A triangle is formed by connecting the points $(0,0), (1,0), \textup{ and }(1,6)$ . Determine the elevation angle to the nearest integer in degrees.

Answer

After connecting the points on the graph, the length of the triangular base is 1 unit.

The height of the triangle is 6. To find the elevation angle, the angle is opposite from the height of the triangle. Since we know the base and the height, the elevation angle can be solved by using the property of tangent.

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

$tan(\theta)= \frac{\textup{6}}{\textup{1}}$

$\theta = tan^{-1}(6)=80.53768 \textup{ degrees}$

The best answer is .

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Question

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.)

Answer

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.

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

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

$c^2 \approx 19358$

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

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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 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.

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Question

In $\bigtriangleup ABC$ :

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

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 $\gamma = m \angle A = 57^{\circ }$ , then evaluating:

$(BC)^{2} = 19 ^{2} + 25 ^{2} - 2 (19)(25)\cos 57^{\circ }$

$(BC)^{2} \approx 361 + 625 - 950 \cdot 0.5446$

$(BC)^{2} \approx 361 + 625 - 517.4$

$(BC)^{2} \approx 468.6$

Taking the square root of both sides:

$BC \approx \sqrt{468.6} \approx 21.6$ .

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Question

In $\bigtriangleup ABC$ :

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

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

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

Answer

The figure referenced is below:

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:

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

$b \approx 49$

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Question

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Answer

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

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Question

Find the length of side .

Answer

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

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Question

Find the area of the triangle.

Answer

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

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Question

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Answer

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Answer

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

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Question

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Answer

Since $\csc\theta=\frac{1}{\sin\theta}$ :

$\frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

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Question

If $\csc \theta < 0$ and $\tan \theta > 0$ , then which of the following must be true about $\theta$ .

Answer

Since cosecant is negative, theta must be in quadrant III or IV.

Since tangent is positive, it must be in quadrant I or III.

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between $\pi$ and $\frac{3\pi}{2}$ .

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Question

If $\cot \alpha = 5$ and $\sin \alpha < 0$ , what is the value of $\sec \alpha$ ?

Answer

Since cotangent is positive and sine is negative, alpha must be in quadrant III. $\cot \alpha = \frac{x}{y}$ then implies that is a point on the terminal side of alpha.

$r=\sqrt{(x^2+y^2)}=\sqrt{26}$

$\sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}$

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Question

For the above triangle, what is $\sec (\theta)$ if , and ?

Answer

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = \frac{17}{15} = 1.13$

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Question

For the above triangle, what is $\cot (\theta)$ if , and ?

Answer

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = \frac{24}{18} = 1.33$

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Question

The point lies on the terminal side of an angle in standard position. Find the secant of the angle.

Answer

Secant is defined to be the ratio of to where is the distance from the origin.

The Pythagoreanr Triple 5, 12, 13 helps us realize that .

Since , the answer is $\frac{13}{12}$ .

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Question

Given angles and in quadrant I, and given,

$\sin x = \frac{3}{5}$ and $\cos y = \frac {5}{13}$ ,

find the value of $\csc (x+y)$ .

Answer

Use the following trigonometric identity to solve this problem.

$\csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}$

Using the Pythagorean triple 3,4,5, it is easy to find $\cos x = \frac {4}{5}$ .

Using the Pythagorean triple 5,12,13, it is easy to find $\sin y = \frac {12}{13}$ .

So substituting all four values into the top equation, we get

$\csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}$

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Question

Evaluate:

Answer

Evaluate each term separately.

$sec(60)= \frac{1}{cos(60)}=\frac{1}{0.5}= 2$

$csc(45)= \frac{1}{sin(45)}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}=\sqrt2$

$sec(60)-csc(45)=2-\frac{2\sqrt2}{2}=2-\sqrt2$

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