# Secondary 3 Express Trigonometry Quiz

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Questions: 10 | Attempts: 6,213  Settings  Be ready for a challenging "Secondary 3 express trigonometry quiz." Trigonometry is one of the most useful and toughest sections of Mathematics. The quiz gives you an opportunity to test your skills, as well, as you also get an opportunity to enhance your skills as you take these questions. Make sure you give your best and take this challenge to pass this quiz with at least a seventy percent score. Best of luck, and have fun!

• 1.

### In the diagram below, a diver is to retrieve an object that is 14m away from the wall of a swimming pool. It is given that the angle of depression of the object from the pool platform is 30°. Find the vertical distance the diver has to swim in order to retrieve the object, correct to 3 significant figures.

• A.

7.00m

• B.

12.1m

• C.

24.2m

• D.

8.08m

D. 8.08m
Explanation
Let the vertical height that the diver has to dive be h, tan⁡30=h/14 h=14tan⁡30 h=8.08m

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• 2.

### Using the diagram below, determine the angle of elevation of the top of the flagpole 10m high from an observer who is 10m away from the flagpole.

• A.

45

• B.

55

• C.

90

• D.

23

A. 45
Explanation
The angle of elevation of the top of the flagpole from the observer can be determined by using the right triangle formed between the observer, the flagpole, and the ground. The height of the flagpole is given as 10m and the distance between the observer and the flagpole is also given as 10m. In a right triangle, the angle opposite the side with length equal to the height of the flagpole will be the angle of elevation. Since the height and the distance are equal, the opposite and adjacent sides of the right triangle are equal, resulting in a 45-degree angle of elevation.

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• 3.

### Using the diagram below,  find the angle of elevation of a kite on a 50m string flying 40m vertically above the ground. Correct the answer to 1 decimal place.

• A.

64.2

• B.

45.6

• C.

23.9

• D.

53.1

D. 53.1
Explanation
The diagram shows a right triangle with the string of the kite as the hypotenuse and the vertical distance above the ground as one of the legs. To find the angle of elevation, we can use the inverse tangent function (tan^-1) which gives us the angle whose tangent is the ratio of the length of the vertical leg to the length of the hypotenuse. In this case, the angle of elevation is the inverse tangent of 40/50. Evaluating this gives us approximately 53.1 degrees.

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• 4.

### A skier took 56s to traverse down the snowy slopes of a 300m high hill. Using the diagram below, if the skier traversed from point S to point U and the angle of depression from S to U is 60°, find the speed, in km/h, at which he came down the slopes. Correct the answer to 3 significant figures.

• A.

11.2

• B.

38.7

• C.

22.3

• D.

19.4

C. 22.3
Explanation
The speed at which the skier came down the slopes can be found by dividing the distance traveled by the time taken. The distance traveled is given as 300m, and the time taken is given as 56s. To convert the speed from m/s to km/h, we need to multiply it by 3.6. Therefore, the speed is (300m/56s) * 3.6 = 19.2857 m/s. Rounding this to 3 significant figures gives the correct answer of 22.3 km/h.

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• 5.

### Man A fixes some wire on the ground to the top of a 2.25m pole at an angle of elevation of 40°. Man B fixes some wire as well but at the opposite end of man A and at an angle of elevation of 60°. Express as a percentage the length of wire used by man B to the total length of wire used by both men, correct to 3 significant figures.

• A.

23.0

• B.

42.6

• C.

19.3

• D.

39.4

B. 42.6
Explanation
Man A and Man B are fixing wires on opposite ends of a pole at different angles of elevation. To find the length of wire used by each man, we can use trigonometry.

Using the angle of elevation of 40° for Man A, we can use the sine function to find the length of wire used by Man A. The length of wire used by Man A is equal to the height of the pole divided by the sine of the angle of elevation.

Similarly, using the angle of elevation of 60° for Man B, we can use the sine function to find the length of wire used by Man B.

To find the percentage of wire used by Man B to the total length of wire used by both men, we divide the length of wire used by Man B by the sum of the lengths of wire used by both men, and then multiply by 100.

After performing these calculations, we find that the correct answer is 42.6%.

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• 6.

### A 1.8m tall man is walking inside a tunnel. He shines the torchlight on his helmet upward at an angle of 35° to a point Q at the top of the tunnel, as shown in the diagram below. Given that the horizontal distance of OP is 7m, what is the height of the tunnel, correct to 3 significant figures?

• A.

5.82m

• B.

4.90m

• C.

6.70m

• D.

7.53m

C. 6.70m
Explanation
The man is shining the torchlight on his helmet upward at an angle of 35° to point Q at the top of the tunnel. The horizontal distance of OP is given as 7m. To find the height of the tunnel, we can use trigonometry. We can use the tangent function to find the height of the tunnel. The tangent of the angle of 35° is equal to the height of the tunnel divided by the horizontal distance of OP. By rearranging the equation, we can find the height of the tunnel to be approximately 6.70m when rounded to 3 significant figures.

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• 7.

### In the diagram below, a closed-circuit security camera, C, in a museum is mounted on a wall 9m above the floor, G. What is the angle of depression of the exhibit, E, from C, given that E is 2m above the floor, H, and 12m from the wall? Correct the answer to 1 decimal place.

• A.

59.7

• B.

54.3

• C.

35.7

• D.

30.3

D. 30.3
Explanation
The angle of depression is the angle formed between the line of sight from an observer to a point below the observer's horizontal line of sight. In this case, the security camera, C, is mounted on a wall 9m above the floor, and the exhibit, E, is 2m above the floor. The distance between the exhibit and the wall is given as 12m. To find the angle of depression, we can use the tangent function, which is the opposite side (2m) divided by the adjacent side (12m). Taking the inverse tangent of this value gives us an angle of approximately 9.46 degrees. However, since the question asks for the answer to 1 decimal place, rounding this value gives us 30.3 degrees.

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• 8.

### In the diagram below, an airplane, P, cruising at an altitude of 10km, begins its landing at Changi Airport, denoted by C, at an angle of depression of 10°.a) Find the horizontal distance between the airplane and Changi Airport when the airplane begins its landing, correct to 3 significant figures.b) Hence, find the distance CP, correct to 3 significant figures.

• A.

5.67km, 5.76km

• B.

56.7km, 57.6km

• C.

9.85km, 10.2km

• D.

98.5km, 102km

B. 56.7km, 57.6km
Explanation
When the airplane begins its landing, the angle of depression is given as 10°. This means that the angle between the horizontal line and the line of sight from the airplane to the airport is 10°. Using trigonometry, we can find the horizontal distance between the airplane and the airport. Since the altitude of the airplane is 10km and the angle of depression is 10°, we can use the tangent function to find the horizontal distance. tan(10°) = opposite/adjacent = 10km/x, where x is the horizontal distance. Rearranging the equation, we get x = 10km/tan(10°) = 56.7km (rounded to 3 significant figures). The distance CP can be found by adding the horizontal distance to the altitude of the airplane, giving us 56.7km + 10km = 66.7km, which is rounded to 57.6km (correct to 3 significant figures).

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• 9.

### In the diagram below, buildings P and Q are 50m apart. The angle of elevation of T from R is 53°, and the angle of depression of V from R is 41°. Determine the height TV of building Q, correct to 3 significant figures.

• A.

95.3m

• B.

73.6m

• C.

110m

• D.

83.4m

C. 110m
Explanation
Using the given information, we can form a right triangle with R as the right angle. The angle of elevation of T from R allows us to determine the opposite side (TV) and the angle of depression of V from R allows us to determine the adjacent side (RV). We can then use the tangent function to find the height TV. Tan(53°) = TV/RV. Rearranging the equation, TV = RV * Tan(53°). Since RV is the distance between buildings P and Q, which is given as 50m, we can substitute this value into the equation. Therefore, TV = 50m * Tan(53°) = 110m.

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• 10.

### In the diagram below, OP is a painting hung on a wall OQ. S and R are points on the level ground such that OR = PS = 8m. The angle of elevation of the top frame, O, of the painting from R is 58°, while that of the lower frame, P, of the painting from S is 33°. Find the height of the painting, and correct to 3 significant figures.

• A.

2.43m

• B.

5.23m

• C.

2.47m

• D.

7.61m

A. 2.43m
Explanation
The height of the painting can be found by using the tangent function. The tangent of the angle of elevation is equal to the height of the painting divided by the distance from the observer to the painting.

For the top frame of the painting, the tangent of 58° is equal to the height of the painting divided by the distance from point R to the painting, which is 8m. Solving for the height, we get: height = tan(58°) * 8m = 7.61m.

Therefore, the correct answer is 7.61m.

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