A right-angled triangle PQR has the right angle at R. Angle P = 40 degrees and the hypotenuse PQ = 12 cm.
(a)Name the side opposite angle P.(1)
(b)Name the side adjacent to angle P (not the hypotenuse).(1)
2
In a right-angled triangle you know an acute angle and the hypotenuse, and you want to find the side adjacent to that angle.
(a)Name the trigonometric ratio you would use.(1)
(b)Write the formula for this ratio in words.(1)
3
Write down, in words, the formula for the tangent ratio in a right-angled triangle.
(1)
4
Write down the exact value of sin(30 degrees).
(1)
5
A right-angled triangle has an acute angle of 35 degrees and hypotenuse 8 cm. Calculate the length of the side opposite this angle, to 1 decimal place.
(2)
6
A right-angled triangle has an acute angle of 52 degrees and hypotenuse 15 cm. Calculate the length of the side adjacent to this angle, to 1 decimal place.
(2)
7
Write down the exact value of cos(60 degrees).
(1)
8
A right-angled triangle has an acute angle of 24 degrees and the side opposite this angle is 6 cm. Find the length of the hypotenuse, to 1 decimal place.
(2)
9
A right-angled triangle has an opposite side of 9 cm and hypotenuse 20 cm. Find the size of the acute angle between them, to the nearest degree.
(2)
10
Write down the exact value of tan(45 degrees).
(1)
11
In a right-angled triangle you know the length of the side opposite angle Y and the length of the hypotenuse, and you want to find angle Y. Which trigonometric ratio should you use?
(1)
12
In a right-angled triangle you know the lengths of the side opposite angle Z and the side adjacent to angle Z, and you want to find angle Z. Which trigonometric ratio should you use?
(1)
13
A ladder rests against a wall, making an angle of 68 degrees with the ground. The foot of the ladder is 3.2 m from the wall. Find the length of the ladder, to 1 decimal place.
(3)
14
A triangle has sides of length 7 cm, 24 cm and 25 cm.
(a)Show that this triangle is right-angled.(1)
(b)Given that the right angle is between the sides of length 7 cm and 24 cm, find the size of the angle opposite the 7 cm side, to the nearest degree.(3)
15
A wheelchair ramp rises 4 m over a horizontal distance of 9 m.
(a)Find the angle of elevation of the ramp, to the nearest degree.(2)
(b)Building regulations state a ramp must not be steeper than 30 degrees. Does this ramp meet the regulation?(1)
16
A right-angled triangle has sides 8 cm (opposite angle X), 15 cm (adjacent to X) and hypotenuse 17 cm.
(a)State the trigonometric ratio equal to 8/17 for angle X.(1)
(b)Use that ratio to find angle X, to the nearest degree.(3)
17
Stretch question. A right-angled triangle has hypotenuse 20 cm. One acute angle is four times the size of the other.
(a)Find the size of each acute angle.(2)
(b)Using the smaller angle, find the length of the side opposite it, to 1 decimal place.(2)
18
Stretch question. A right-angled triangle has one acute angle of 27 degrees and the side adjacent to it measures 9 cm. The opposite side is then increased by 15% while the adjacent side stays the same.
(a)Find the length of the original opposite side, to 2 decimal places.(1)
(b)Find the new opposite side length after the 15% increase, to 2 decimal places.(1)
(c)Calculate the new angle, to the nearest degree.(2)
Mark scheme · KS3.M-G15D Introduction to Trigonometry: Fluency and Exam Drill
Question 1
(a) B1 QR named as the side opposite angle P
(a) Answer: QR
(b) B1 PR named as the side adjacent to angle P
(b) Answer: PR
Question 2
(a) B1 cosine named
(a) Answer: Cosine
(b) B1 cosine = adjacent divided by hypotenuse stated
(b) Answer: Cosine = adjacent / hypotenuse
Question 3
B1 tangent = opposite divided by adjacent stated
Answer: Tangent = opposite / adjacent
Question 4
B1 0.5 or 1/2 cao
Answer: 0.5
Question 5
M1 method: opposite = sin(35) x 8
A1 4.6 cm to 1 dp cao
Answer: 4.6 cm
Question 6
M1 method: adjacent = cos(52) x 15
A1 9.2 cm to 1 dp cao
Answer: 9.2 cm
Question 7
B1 0.5 or 1/2 cao
Answer: 0.5
Question 8
M1 method: hypotenuse = 6 / sin(24)
A1 14.8 cm to 1 dp cao
Answer: 14.8 cm
Question 9
M1 method: angle = sin^{-1}(9/20)
A1 27 degrees to nearest degree cao
Answer: 27 degrees
Question 10
B1 1 cao
Answer: 1
Question 11
B1 sine cao
Answer: Sine
Question 12
B1 tangent cao
Answer: Tangent
Question 13
M1 method: ladder length = 3.2 / cos(68)
A1 substitution shown: 3.2 / cos(68)
A1 8.5 m to 1 dp cao
Answer: 8.5 m
Question 14
(a) B1 calculation 72 + 242 = 625 = 252 shown
(a) Answer: 49 + 576 = 625 = 252
(b) M1 method: angle = sin^{-1}(7/25) or equivalent
(b) A1 correct substitution and evaluation shown
(b) A1 16 degrees to nearest degree cao
(b) Answer: 16 degrees
Question 15
(a) M1 method: angle = tan^{-1}(4/9)
(a) A1 24 degrees to nearest degree cao
(a) Answer: 24 degrees
(b) B1 yes, because 24 degrees is less than 30 degrees, ft from part (a)
(b) Answer: Yes, because 24 degrees is less than 30 degrees
Question 16
(a) B1 sine named since opposite/hypotenuse = 8/17
(a) Answer: Sine
(b) M1 method: X = sin^{-1}(8/17)
(b) A1 correct substitution and evaluation shown
(b) A1 28 degrees to nearest degree cao
(b) Answer: 28 degrees
Question 17
(a) M1 method: recognise angles sum to 90, so x + 4x = 90, hence x = 18
(a) A1 angles are 18 degrees and 72 degrees cao
(a) Answer: 18 degrees and 72 degrees
(b) M1 method: opposite = 20 x sin(18)
(b) A1 6.2 cm to 1 dp cao
(b) Answer: 6.2 cm
Question 18
(a) B1 4.59 cm cao
(a) Answer: 4.59 cm
(b) B1 5.27 cm cao ft
(b) Answer: 5.27 cm
(c) M1 method: new angle = tan^{-1}(new opposite / 9), ft from part (b)