Math 2 Quiz: Using Trig Ratios For Sides

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Using Trig Ratios For SidesQuestion 1 of 14

A right circular cone has base radius $6$ and slant height $10$ . A plane parallel to the base intersects the cone creating a circular cross-section. If this cross-section is $\frac{3}{4}$ of the way up from the base to the apex, what is the radius of this circular cross-section?

4.5

units

2.25

units

1.5

units

\frac{15}{8}

units

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Math 2 Quiz

Math 2 Quiz: Using Trig Ratios For Sides

Practice Using Trig Ratios For Sides in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Trig Ratios For Sides, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$4.5$ units
$2.25$ units
$1.5$ units (correct answer)
$\frac{15}{8}$ units

Explanation: When you encounter a cone cross-section problem, you're dealing with similar triangles. The key insight is that when a plane cuts a cone parallel to its base, the resulting cross-section creates a smaller cone that's geometrically similar to the original. First, find the cone's height using the Pythagorean theorem. With base radius 6 and slant height 10:

h^2 + 6^2 = 10^2

, so

h^2 = 100 - 36 = 64

, giving us

h = 8

. Since the cross-section is

\frac{3}{4}

of the way up from base to apex, it's at height

\frac{3}{4} \times 8 = 6

from the base, or 2 units down from the apex. The small cone above the cross-section has height 2. Using similar triangles, the ratio of corresponding parts is constant:

\frac{\text{small cone height}}{\text{original height}} = \frac{\text{small radius}}{\text{original radius}}

So:

\frac{2}{8} = \frac{r}{6}

, which gives us

r = 6 \times \frac{1}{4} = 1.5

. Answer A)

4.5

incorrectly uses

\frac{3}{4} \times 6 = 4.5

, confusing the position fraction with the similarity ratio. Answer B)

2.25

likely comes from

\frac{3}{8} \times 6

, mixing up height relationships. Answer D)

\frac{15}{8}

appears to involve incorrect fraction manipulation. Remember: in similar triangle problems involving cones, always work with the distance from the apex, not from the base. The similarity ratio equals the fraction of the total height measured from the apex.

Question 2

A right triangular prism has a right triangular base where one acute angle measures $35°$ and the hypotenuse of the base is $20$ cm. If the prism's height is equal to the shorter leg of the triangular base, what is the volume of the prism?

$2000\sin(35°)\cos(35°)\sin(55°)$ cubic cm
$2000\sin^2(35°)\cos(35°)$ cubic cm (correct answer)
$2000\sin(35°)\cos^2(35°)$ cubic cm
$2000\sin(35°)\cos(35°)\tan(35°)$ cubic cm

Explanation: In the right triangle base with hypotenuse 20 cm and acute angle 35°, the legs are

20\sin(35°)

and

20\cos(35°)

. The shorter leg is

20\sin(35°)

(since

35° < 45°

, so

\sin(35°) < \cos(35°)

). The base area is

\frac{1}{2} \cdot 20\sin(35°) \cdot 20\cos(35°) = 200\sin(35°)\cos(35°)

. The volume is base area × height =

200\sin(35°)\cos(35°) \cdot 20\sin(35°) = 2000\sin^2(35°)\cos(35°)

. Choice A incorrectly uses

\sin(55°)

instead of

\sin(35°)

for the height. Choice C assumes the longer leg is the height. Choice D uses

\tan(35°)

instead of

\sin(35°)

for the height calculation.

Question 3

Two observers are $500$ meters apart on level ground. They simultaneously measure the angles of elevation to an airplane directly above the line connecting them. The first observer measures $72°$ and the second measures $81°$ . How high is the airplane above the ground?

$\frac{500\sin(72°)\sin(81°)}{\sin(81°) - \sin(72°)}$ meters
$\frac{500\tan(72°)\tan(81°)}{\tan(81°) - \tan(72°)}$ meters
$\frac{500\tan(72°)\tan(81°)}{\tan(72°) + \tan(81°)}$ meters (correct answer)
$500\tan(72°)\cos(81°)$ meters

Explanation: When you encounter angle of elevation problems with two observers, you're dealing with a classic trigonometry setup that requires careful coordinate analysis. The key insight is recognizing that the airplane creates two right triangles sharing the same height. Let's set up coordinates with the first observer at the origin and the second at position 500. If the airplane is at horizontal distance

x

from the first observer and height

h

, then:

From the first observer: $\tan(72°) = \frac{h}{x}$ , so $h = x\tan(72°)$
From the second observer: $\tan(81°) = \frac{h}{500-x}$ , so $h = (500-x)\tan(81°)$

Setting these equal:

x\tan(72°) = (500-x)\tan(81°)

Expanding:

x\tan(72°) = 500\tan(81°) - x\tan(81°)

Collecting

x

terms:

x(\tan(72°) + \tan(81°)) = 500\tan(81°)

Therefore:

x = \frac{500\tan(81°)}{\tan(72°) + \tan(81°)}

The height is:

h = x\tan(72°) = \frac{500\tan(81°)\tan(72°)}{\tan(72°) + \tan(81°)}

This matches answer choice C. Answer A incorrectly uses sine functions instead of tangent for angle of elevation problems. Answer B has the wrong denominator (subtraction instead of addition), which would arise from incorrectly setting up the distance relationships. Answer D oversimplifies by ignoring the second observer entirely. Remember: angle of elevation problems always use tangent (opposite over adjacent), and when you have two observers, the distances to the object must sum correctly to the distance between observers.

Question 4

A kite is flying at the end of a $150$ -meter string. The string makes a $62°$ angle with the horizontal ground. Due to wind, the kite moves horizontally $20$ meters while the string length remains constant. What is the kite's new height above the ground?

$150\sin(62°) - 20\tan(62°)$ meters
$\sqrt{(150\sin(62°))^2 + (150\cos(62°) - 20)^2} - 150\cos(62°)$ meters
$150\sin(62°) - \sqrt{400 - (150\cos(62°))^2}$ meters
$\sqrt{150^2 - (150\cos(62°) + 20)^2}$ meters (correct answer)

Explanation: Initially, the kite is at height

h_1 = 150\sin(62°)

and horizontal distance

d_1 = 150\cos(62°)

from the person. After moving

20

meters horizontally, the new horizontal distance is

d_2 = 150\cos(62°) + 20

. Since the string length remains

150

meters, using the Pythagorean theorem:

h_2^2 + d_2^2 = 150^2

. Therefore,

h_2 = \sqrt{150^2 - d_2^2} = \sqrt{150^2 - (150\cos(62°) + 20)^2}

. Choice A incorrectly subtracts a tangent term. Choice B attempts to use the distance formula but sets up the calculation incorrectly. Choice C assumes the kite moves vertically as well, which isn't stated in the problem.

Question 5

A ladder leans against a vertical wall forming a 65° angle with the horizontal ground. If the foot of the ladder is 8 feet from the base of the wall, what is the length of the ladder to the nearest foot?

19 feet (correct answer)
9 feet
17 feet
7 feet

Explanation: The ladder forms the hypotenuse of a right triangle. The horizontal distance (8 feet) is adjacent to the 65° angle. Using cos(65°) = adjacent/hypotenuse = 8/ladder length, we get ladder length = 8/cos(65°) = 8/0.4226 ≈ 19 feet. Choice B uses tan(65°) × 8. Choice C uses sin(65°) × 8. Choice D uses 8 × cos(65°).

Question 6

A kite string makes a 58° angle with the horizontal ground. If 85 meters of string have been let out and the string is taut, what is the vertical height of the kite above the ground to the nearest meter?

45 meters
72 meters (correct answer)
136 meters
100 meters

Explanation: When you encounter a problem involving angles and distances, you're likely dealing with trigonometry. This scenario creates a right triangle where the kite string is the hypotenuse, the ground is the horizontal leg, and the kite's height is the vertical leg. To find the vertical height, you need to identify which trigonometric function relates the given information. You know the hypotenuse (85 meters) and the angle with the horizontal (58°), and you want the opposite side (height). The sine function connects these:

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}

So:

\sin(58°) = \frac{\text{height}}{85}

Solving for height:

\text{height} = 85 \times \sin(58°) = 85 \times 0.848 = 72.08

meters Rounded to the nearest meter, this gives you 72 meters, which is answer B. Looking at the wrong answers: A) 45 meters results from using cosine instead of sine, giving you the horizontal distance rather than the vertical height. C) 136 meters likely comes from incorrectly using

\frac{85}{\sin(58°)}

, which would give the hypotenuse if you knew the opposite side. D) 100 meters doesn't correspond to any standard trigonometric calculation with these values. Remember the SOH-CAH-TOA mnemonic: Sine = Opposite/Hypotenuse. When you have the hypotenuse and need the side opposite to your given angle, sine is your tool. Always double-check which side you're solving for relative to your angle.

Question 7

A surveyor measures the angle of elevation to the top of a building as 38° from a point 120 meters from the base. If the surveyor's instrument is 1.5 meters above ground level, what is the total height of the building to the nearest meter?

96 meters
94 meters
75 meters
95 meters (correct answer)

Explanation: When you encounter angle of elevation problems, you're dealing with right triangle trigonometry where you need to carefully account for all height measurements involved. Here, you have a right triangle where the horizontal distance is 120 meters, the angle of elevation is 38°, and you need the total building height. The key insight is that the surveyor's instrument creates two separate height components. Using trigonometry, the height from the instrument level to the building top is:

\tan(38°) \times 120 = 0.7813 \times 120 = 93.8

meters. Since the instrument is 1.5 meters above ground, the total building height is

93.8 + 1.5 = 95.3

meters, which rounds to 95 meters. Choice A (96 meters) likely comes from rounding 93.8 up to 94, then adding 1.5 to get 95.5, which rounds to 96. This involves incorrect intermediate rounding. Choice B (94 meters) represents forgetting to add the instrument height entirely—just rounding 93.8 to 94. Choice C (75 meters) suggests using the wrong trigonometric ratio, possibly

\sin(38°) \times 120 = 73.9

, then adding 1.5. The correct answer is D (95 meters). Remember for elevation problems: always identify what the angle is measured from (instrument level, not ground level), use the appropriate trig function for what you're solving, and account for all height offsets. Don't round intermediate calculations—only round your final answer.

Question 8

A ramp is designed to rise 3.5 feet over a horizontal distance of 28 feet. If a support beam is placed perpendicular to the ramp surface, connecting the highest point of the ramp to the ground directly below it, what angle does the ramp make with the horizontal ground to the nearest degree?

83°
7° (correct answer)
14°
76°

Explanation: When you encounter a ramp problem like this, you're dealing with right triangle trigonometry. The ramp creates the hypotenuse of a right triangle, with the vertical rise and horizontal distance forming the two legs. To find the angle the ramp makes with the horizontal ground, you need to identify which trigonometric ratio to use. You have the opposite side (vertical rise = 3.5 feet) and the adjacent side (horizontal distance = 28 feet) relative to the angle you're seeking. This means you should use the tangent ratio:

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3.5}{28} = 0.125

To find the angle, take the inverse tangent:

\theta = \tan^{-1}(0.125) \approx 7.1°

, which rounds to 7°. Looking at the wrong answers: A) 83° would be the complementary angle (90° - 7°), which represents the steep angle the ramp makes with the vertical, not the horizontal. C) 14° is approximately double the correct answer, possibly resulting from a calculation error or using the wrong ratio. D) 76° is another steep angle that doesn't match the gentle slope described by a rise of 3.5 feet over 28 feet. The correct answer is B) 7°. Remember that gentle ramps have small angles with the horizontal ground. When the horizontal distance is much larger than the vertical rise, expect a small angle. Always double-check that your answer makes physical sense with the problem description.

Question 9

In a right triangle, one leg is $\sqrt{3}$ times the length of the other leg. If the perimeter of the triangle is $12 + 4\sqrt{3}$ , what is the length of the hypotenuse?

$8$ units (correct answer)
$6 + 2\sqrt{3}$ units
$4 + 4\sqrt{3}$ units
$6\sqrt{3}$ units

Explanation: Let the legs be

x

and

x\sqrt{3}

. By the Pythagorean theorem, the hypotenuse is

\sqrt{x^2 + (x\sqrt{3})^2} = \sqrt{x^2 + 3x^2} = 2x

. The perimeter is

x + x\sqrt{3} + 2x = x(3 + \sqrt{3}) = 12 + 4\sqrt{3}

. Factoring the right side:

12 + 4\sqrt{3} = 4(3 + \sqrt{3})

. Therefore,

x(3 + \sqrt{3}) = 4(3 + \sqrt{3})

, so

x = 4

. The hypotenuse is

2x = 8

. Choice B represents one of the legs plus part of the other. Choice C incorrectly adds the perimeter components. Choice D uses

x\sqrt{3}

instead of

2x

for the hypotenuse.

Question 10

A regular hexagon is inscribed in a circle of radius $8$ . What is the length of the apothem (perpendicular distance from center to any side) of this hexagon?

$4\sqrt{3}$ units (correct answer)
$4\sqrt{3}$ units
$4$ units
$4\sqrt{2}$ units

Explanation: A regular hexagon inscribed in a circle can be divided into 6 equilateral triangles, each with a central angle of

60°

. The apothem is the distance from the center to the midpoint of any side. Consider the right triangle formed by the center, a vertex of the hexagon, and the midpoint of an adjacent side. The hypotenuse is the radius (8), and the angle at the center is

30°

(half of

60°

). The apothem is the adjacent side to this

30°

angle, so: apothem =

8\cos(30°) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}

. Choice B would result from calculation errors. Choice C gives

8\sin(30°) = 4

, which is half the side length, not the apothem. Choice D uses

\sqrt{2}

instead of

\sqrt{3}

, suggesting confusion with a square.

Question 11

In triangle ABC, angle C is a right angle. If the length of side BC is 15 units and the measure of angle A is 28°, what is the length of side AC to the nearest tenth of a unit?

28.2 units (correct answer)
7.1 units
8.0 units
13.2 units

Explanation: In this right triangle, BC (opposite to angle A) = 15 and we need AC (adjacent to angle A). Using tan(28°) = opposite/adjacent = BC/AC, we get tan(28°) = 15/AC. Solving: AC = 15/tan(28°) = 15/0.5317 ≈ 28.2 units. Choice B uses sin instead of tan. Choice C uses cos(28°) × 15. Choice D uses sin(28°) × 15.

Question 12

In triangle DEF with right angle at E, if DE = 9.2 units and angle F = 35°, what is the length of EF to the nearest tenth of a unit?

6.6 units
5.3 units
13.1 units (correct answer)
11.3 units

Explanation: When you encounter a right triangle problem with one side and one acute angle given, trigonometric ratios are your key tool. You need to identify which sides are opposite, adjacent, and hypotenuse relative to the given angle. In triangle DEF with the right angle at E, you have DE = 9.2 units and angle F = 35°. From angle F's perspective: EF is the adjacent side, DE is the opposite side, and DF is the hypotenuse. Since you know the opposite side and need the adjacent side, use the tangent ratio:

\tan(35°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{DE}{EF} = \frac{9.2}{EF}

Solving for EF:

EF = \frac{9.2}{\tan(35°)} = \frac{9.2}{0.7002} ≈ 13.1

units. Looking at the wrong answers: Choice A (6.6 units) results from incorrectly multiplying 9.2 by sin(35°) instead of dividing by tan(35°). Choice B (5.3 units) comes from multiplying 9.2 by cos(35°), which would give you a side length if you were using angle D instead of angle F. Choice D (11.3 units) appears to use an incorrect trigonometric calculation, possibly confusing the setup. The correct answer is C) 13.1 units. Study tip: Always draw and label your triangle first, then identify which trigonometric ratio connects your known and unknown values. Remember: SOH-CAH-TOA helps you choose the right ratio, and when you need to "undo" a trig function, use division rather than multiplication.

Question 13

In right triangle PQR with right angle at Q, if PR = 24 cm and angle P = 42°, what is the length of QR to the nearest centimeter?

14 cm
18 cm
36 cm
16 cm (correct answer)

Explanation: When you encounter a right triangle problem with one angle and the hypotenuse given, trigonometric ratios are your key tools. You need to identify which side you're looking for relative to the given angle to choose the correct ratio. In this triangle, you have the hypotenuse PR = 24 cm, angle P = 42°, and you need to find QR. Since QR is the side opposite to angle P, you'll use the sine ratio:

\sin(P) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{QR}{PR}

Substituting the known values:

\sin(42°) = \frac{QR}{24}

Solving for QR:

QR = 24 \times \sin(42°) = 24 \times 0.6691 = 16.06

cm Rounded to the nearest centimeter, QR = 16 cm, making D correct. Let's examine why the other answers are wrong. Choice A (14 cm) would result from using an incorrect sine value or making a calculation error. Choice B (18 cm) might come from using cosine instead of sine:

24 \times \cos(42°) ≈ 24 \times 0.74 = 18

cm, but cosine gives you the adjacent side (PQ), not the opposite side (QR). Choice C (36 cm) is impossible since no side in a right triangle can be longer than the hypotenuse. Remember the mnemonic SOH-CAH-TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. Always identify which side you need relative to your given angle before choosing your trigonometric ratio.

Question 14

A cable car travels up a mountain at a constant 22° angle of inclination. If the cable car travels 450 meters along the cable, what is the horizontal distance covered to the nearest meter?

485 meters
168 meters
417 meters (correct answer)
179 meters

Explanation: When you see a problem involving an inclined path and need to find horizontal distance, you're working with right triangle trigonometry. The cable car creates the hypotenuse of a right triangle, where the horizontal distance is the adjacent side to the 22° angle. To find the horizontal distance, you need the cosine function:

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}

. Here,

\cos(22°) = \frac{\text{horizontal distance}}{450}

. Solving for horizontal distance:

\text{horizontal distance} = 450 \times \cos(22°) = 450 \times 0.9272 = 417.24

meters, which rounds to 417 meters. Choice A (485 meters) represents a common error where students might have used

450 \div \cos(22°)

instead of multiplying, or confused the relationship entirely. Choice B (168 meters) comes from incorrectly using sine instead of cosine:

450 \times \sin(22°) = 450 \times 0.3746 = 168.57

meters. This would give you the vertical height, not horizontal distance. Choice D (179 meters) might result from using the wrong angle or making a calculation error with the trigonometric functions. Remember the mnemonic SOH-CAH-TOA: for horizontal distance on an incline, you want the adjacent side, so use cosine (CAH). When the angle and hypotenuse are given, multiply by cosine to find the horizontal component. Always double-check whether the problem asks for horizontal or vertical distance, as mixing up sine and cosine is a frequent trap.