Junior Certificate Ordinary Level Trigonometry
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Junior Certificate Ordinary Level Trigonometry Question 1 Junior Certificate Ordinary Level (ii) Write down, as a fraction, the value of sin A. (a) The right-angled triangle in the diagram has measurements as shown. A 12 (i) Write down the
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01
Junior Certificate
Ordinary Level
Trigonometry<br>
Ordinary Level
Trigonometry<br>
02
Question 1 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
03
(ii) Write down, as a fraction,
the value of sin A. (a) The right-angled triangle in the diagram has measurements
as shown. A 12 (i) Write down the length of the side opposite to angle A. 15 Length of the side opposite to angle A = 9 15 HYP OPP 12 12 sin A =<br>
the value of sin A. (a) The right-angled triangle in the diagram has measurements
as shown. A 12 (i) Write down the length of the side opposite to angle A. 15 Length of the side opposite to angle A = 9 15 HYP OPP 12 12 sin A =<br>
04
–––––––––– 13 (ii) Using the diagram, complete
the following (iii) Hence calculate | XY |. hypotenuse (b) In the right-angled triangle XYZ,
| XZ | = 13, | ∠ YXZ | = 60º. (i) Using your calculator, write
down the value of cos 60º. cos 60º = 0·5 ADJ adjacent 13 X HYP 60° Y Z | XY | cos 60° = –––– | XY | = 13 × 0·5 = 6·5<br>
the following (iii) Hence calculate | XY |. hypotenuse (b) In the right-angled triangle XYZ,
| XZ | = 13, | ∠ YXZ | = 60º. (i) Using your calculator, write
down the value of cos 60º. cos 60º = 0·5 ADJ adjacent 13 X HYP 60° Y Z | XY | cos 60° = –––– | XY | = 13 × 0·5 = 6·5<br>
05
1·917… (c) As part of an activity lesson a group of students was
asked to measure the height of the mast [RS ].
The mast, [RS ], is supported by the cable [RT ]. S T R 10 m 50° OPP (i) Find the height of the mast RS.
Give your answer correct to the nearest metre. h The students measured the distance from
S to T and they also measured the angle ∠ STR.
They found ST = 10 m and | ∠ STR | = 50°. ADJ 10 tan 50° = | RS | tan 50° = | RS | = 1 2 m 12 m<br>
asked to measure the height of the mast [RS ].
The mast, [RS ], is supported by the cable [RT ]. S T R 10 m 50° OPP (i) Find the height of the mast RS.
Give your answer correct to the nearest metre. h The students measured the distance from
S to T and they also measured the angle ∠ STR.
They found ST = 10 m and | ∠ STR | = 50°. ADJ 10 tan 50° = | RS | tan 50° = | RS | = 1 2 m 12 m<br>
06
12 m (c) As part of an activity lesson a group of students was
asked to measure the height of the mast [RS ].
The mast, [RS ], is supported by the cable [RT ]. S T R 10 m 50° c (ii) Using the theorem of Pythagoras, or otherwise, find the
length of the supporting cable, RT .
Give your answer correct to the nearest metre. The students measured the distance from
S to T and they also measured the angle ∠ STR.
They found ST = 10 m and | ∠ STR | = 50°. 100 + 144 244 Pythagoras’ Theorem a2 + b2 = c2 c2 = 102 + 122 c 2 = c = 1 5·62… 6 m<br>
asked to measure the height of the mast [RS ].
The mast, [RS ], is supported by the cable [RT ]. S T R 10 m 50° c (ii) Using the theorem of Pythagoras, or otherwise, find the
length of the supporting cable, RT .
Give your answer correct to the nearest metre. The students measured the distance from
S to T and they also measured the angle ∠ STR.
They found ST = 10 m and | ∠ STR | = 50°. 100 + 144 244 Pythagoras’ Theorem a2 + b2 = c2 c2 = 102 + 122 c 2 = c = 1 5·62… 6 m<br>
07
Question 2 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
08
41 9 (a) The diagram shows a right-angled triangle with measurements
as shown. HYP ADJ X 9 41 (i) Write down the length of the hypotenuse of the triangle. (ii) Write down the value of cos X as a fraction. 40<br>
as shown. HYP ADJ X 9 41 (i) Write down the length of the hypotenuse of the triangle. (ii) Write down the value of cos X as a fraction. 40<br>
09
(b) In the right-angled triangle ABC,
| AC | = 25
| ∠ BAC | = 32°.
Let | BC | = x. B (i) Using your calculator find sin 32°.
Write your answer correct to two decimal places. 25 HYP OPP 32° A C sin 32° = (ii) Using the diagram of the triangle ABC write sin 32° as a fraction. sin 32° = hypotenuse opposite x 25 –––––––––– 0·5 299… x 3 (iii) Hence, or otherwise, find x, the value of | CB | . = x = 25 × 0·53 = 13·25<br>
| AC | = 25
| ∠ BAC | = 32°.
Let | BC | = x. B (i) Using your calculator find sin 32°.
Write your answer correct to two decimal places. 25 HYP OPP 32° A C sin 32° = (ii) Using the diagram of the triangle ABC write sin 32° as a fraction. sin 32° = hypotenuse opposite x 25 –––––––––– 0·5 299… x 3 (iii) Hence, or otherwise, find x, the value of | CB | . = x = 25 × 0·53 = 13·25<br>
10
(c) Seán wishes to measure the width of a canal.
He is at a point A directly opposite a landmark B on the
opposite bank. Seán walks 50 paces along the bank of the
canal to point C.
He measures the angle ACB and finds it is 25°. B 25° A C (i) Each of Seán’s paces is 0·7 m. Calculate | AC | . 50 × 0·7 = 35 m (ii) Hence calculate the width of the canal, | AB | .
Give your answer to the nearest metre. x 35 m –––––––– adjacent opposite tan 25° = x 35 OPP ADJ x = 35 × tan 25° = 16 ·3207.. m<br>
He is at a point A directly opposite a landmark B on the
opposite bank. Seán walks 50 paces along the bank of the
canal to point C.
He measures the angle ACB and finds it is 25°. B 25° A C (i) Each of Seán’s paces is 0·7 m. Calculate | AC | . 50 × 0·7 = 35 m (ii) Hence calculate the width of the canal, | AB | .
Give your answer to the nearest metre. x 35 m –––––––– adjacent opposite tan 25° = x 35 OPP ADJ x = 35 × tan 25° = 16 ·3207.. m<br>
11
Question 3 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
12
(a) The right-angled triangle ABC has measurements as shown. ADJ OPP C C A B 26 24 10 (i) Write down the length of the side opposite the angle C. 24 (ii) Write down the value of tan C, as a fraction. 24 10 Length of the side opposite to the angle C =<br>
13
(b) In the right-angled triangle PQR,
| PQ | = 14 and | ∠ PQR | = 53º.
Let | QR | = d. P Q R d 53º 14 (i) Using the diagram write down the
value of cos 53°, as a fraction. –––––––––– hypotenuse adjacent cos 53° = HYP ADJ d 14 (ii) Using your calculator, or otherwise, write down the value
of cos 53° correct to one decimal place. cos 53° = 0·6018…. (iii) Hence find d, the value of | QR |. = 0·6 d = 14 × 0·6 = 8·4<br>
| PQ | = 14 and | ∠ PQR | = 53º.
Let | QR | = d. P Q R d 53º 14 (i) Using the diagram write down the
value of cos 53°, as a fraction. –––––––––– hypotenuse adjacent cos 53° = HYP ADJ d 14 (ii) Using your calculator, or otherwise, write down the value
of cos 53° correct to one decimal place. cos 53° = 0·6018…. (iii) Hence find d, the value of | QR |. = 0·6 d = 14 × 0·6 = 8·4<br>
14
| ZW | = 9 m W (c) [ZW] is a vertical television aerial mast.
[ZX] and [ZY] are supporting cables.
| ZX | = 15 m, | XW | = 12 m and |WY | = 16 m. (i) In Δ XWZ, use the Theorem of Pythagoras, to find | ZW | ,
the height of the television aerial mast. Z 15 m 12 m 16 m X Y A Theorem of Pythagoras (Side 1)2 + (Side 2)2 = (Hypotenuse)2 | ZW |2 + 122 = 152 | ZW | = 2 144 + 225 Hypotenuse – 81 9 m<br>
[ZX] and [ZY] are supporting cables.
| ZX | = 15 m, | XW | = 12 m and |WY | = 16 m. (i) In Δ XWZ, use the Theorem of Pythagoras, to find | ZW | ,
the height of the television aerial mast. Z 15 m 12 m 16 m X Y A Theorem of Pythagoras (Side 1)2 + (Side 2)2 = (Hypotenuse)2 | ZW |2 + 122 = 152 | ZW | = 2 144 + 225 Hypotenuse – 81 9 m<br>
15
(ii) Hence find the measure of the angle marked A in the diagram
correct to the nearest degree. 15 m 12 m 16 m A 9 m ADJ OPP 16 9 = 0·5625 A = tan–10·5625 = 29 ·3577… ° W Z X Y (c) [ZW] is a vertical television aerial mast.
[ZX] and [ZY] are supporting cables.
| ZX | = 15 m, | XW | = 12 m and |WY | = 16 m.<br>
correct to the nearest degree. 15 m 12 m 16 m A 9 m ADJ OPP 16 9 = 0·5625 A = tan–10·5625 = 29 ·3577… ° W Z X Y (c) [ZW] is a vertical television aerial mast.
[ZX] and [ZY] are supporting cables.
| ZX | = 15 m, | XW | = 12 m and |WY | = 16 m.<br>
16
Question 4 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
17
(a) The right-angled triangle ABC has measurements as shown. HYP OPP C B B A 17 15 8 (i) Write down the length of the hypotenuse of the Δ ABC. 17 (ii) Write down the value of sin B, as a fraction. 8 17<br>
18
(ii) Using your calculator, or otherwise,
write down the value of tan | ∠ PQR|
correct to one decimal place. (iii) Hence, or otherwise, calculate | PR |
correct to one decimal place. adjacent (b) In the right-angled triangle PQR,
| QR | = 4, |∠ QPR| = 48º and |∠ PQR| = t º. 48º P Q R (i) Find the value of t. tan |∠ PQR| = tan 42º 4 ADJ OPP opposite tan 42º = | PR | 4 0·9 = Multiply by 4 × | PR | = 3·6 t º t º = 90º – 48º = 42º 42 º = 0·9004..<br>
write down the value of tan | ∠ PQR|
correct to one decimal place. (iii) Hence, or otherwise, calculate | PR |
correct to one decimal place. adjacent (b) In the right-angled triangle PQR,
| QR | = 4, |∠ QPR| = 48º and |∠ PQR| = t º. 48º P Q R (i) Find the value of t. tan |∠ PQR| = tan 42º 4 ADJ OPP opposite tan 42º = | PR | 4 0·9 = Multiply by 4 × | PR | = 3·6 t º t º = 90º – 48º = 42º 42 º = 0·9004..<br>
19
242 + | AC |2 = 252 (c) In the Δ ABC, |∠ BCA| = 90º, | AB | = 25 m and | BC | = 24 m. 24 m B C A 25 m Pythagoras’ Theorem (Side 1)2 + (Side 2)2 = (Hypotenuse)2 | AC |2 = 625 576 + – 49 | AC | 2 = = 7 m 7 m (i) Find, in metres, | AC |. = = 7 m Pythagoras’ Theorem<br>
20
24 25 = ___ (ii) Find |∠ BAC |, correct to the nearest degree. 24 m |∠ BAC| = 73·739… 25 m 7 m OPP HYP sin |∠ BAC| = = 74º B C A (c) In the Δ ABC, |∠ BCA| = 90º, | AB | = 25 m and | BC | = 24 m.<br>
21
Question 5 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
22
(a) The right-angled triangle ABC has measurements as shown. ADJ B C C A 6 10 8 (i) Write down the length of the hypotenuse of the Δ ABC. 10 (ii) Write down the value of cos C,
as a fraction. 10 6 HYP<br>
as a fraction. 10 6 HYP<br>
23
(b) In the right-angled triangle PQR,
| PQ| = 8 and |∠ PQR| = 50º. 50º P R Q (i) Find |∠ QPR |. (ii) Using your calculator, or otherwise, write down the value of
sin |∠ QPR | correct to two decimal places. 8 sin 40º = 0·64278….. |∠ QPR | = 180º – 90º – 50º. = 40º 40º<br>
| PQ| = 8 and |∠ PQR| = 50º. 50º P R Q (i) Find |∠ QPR |. (ii) Using your calculator, or otherwise, write down the value of
sin |∠ QPR | correct to two decimal places. 8 sin 40º = 0·64278….. |∠ QPR | = 180º – 90º – 50º. = 40º 40º<br>
24
hypotenuse 50º 5 (iii) Hence, or otherwise, calculate | QR |
correct to one decimal place. 8 OPP HYP opposite sin 40º = | QR| 8 Multiply both sides by 8 | QR| = 8 × 0·64 40º ✍ From part (ii) | QR| = 5·12 | QR| = 5·1 P R Q (b) In the right-angled triangle PQR,
| PQ| = 8 and |∠ PQR| = 50º.<br>
correct to one decimal place. 8 OPP HYP opposite sin 40º = | QR| 8 Multiply both sides by 8 | QR| = 8 × 0·64 40º ✍ From part (ii) | QR| = 5·12 | QR| = 5·1 P R Q (b) In the right-angled triangle PQR,
| PQ| = 8 and |∠ PQR| = 50º.<br>
25
(c) In the Δ ABC, |∠ BCA | = 90º, | AB | =13 m and | AC | = 5 m.
(i) Find, in metres, | BC |. 5 m B C 13 m A Pythagoras’ Theorem (Side 1)2 + (Side 2)2 = (Hypotenuse)2 52 + | BC |2 = 132 | BC |2 = 169 25 + – 144 | BC | 2 = = 12 m ✍<br>
(i) Find, in metres, | BC |. 5 m B C 13 m A Pythagoras’ Theorem (Side 1)2 + (Side 2)2 = (Hypotenuse)2 52 + | BC |2 = 132 | BC |2 = 169 25 + – 144 | BC | 2 = = 12 m ✍<br>
26
A 13 12 (c) In the Δ ABC, |∠ BCA | = 90º, | AB | =13 m and | AC | = 5 m.
(ii) Find |∠ BAC |, correct to the nearest degree. 5 m 13 m 12 m OPP HYP ADJ |∠ BAC | = 67·38.. º B C<br>
(ii) Find |∠ BAC |, correct to the nearest degree. 5 m 13 m 12 m OPP HYP ADJ |∠ BAC | = 67·38.. º B C<br>
27
Question 6 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
28
(a) The right-angled triangle ABC has measurements as shown. ADJ OPP C A B A 24 25 7 (i) Write down the length of the side opposite the angle A. 24 (ii) Write down the value of tan A,
as a fraction. 24 7<br>
as a fraction. 24 7<br>
29
hypotenuse (b) In the right-angled triangle PQR,
| PQ| = 12 and |∠ QPR| = 60º. 60º P Q R (i) Write down the value
of cos 60º. cos 60º = 0·5 (ii) Calculate | PR|. 12 ADJ HYP adjacent cos 60º = | PR| 12 0·5 = Cross multiply × | PR| = 6<br>
| PQ| = 12 and |∠ QPR| = 60º. 60º P Q R (i) Write down the value
of cos 60º. cos 60º = 0·5 (ii) Calculate | PR|. 12 ADJ HYP adjacent cos 60º = | PR| 12 0·5 = Cross multiply × | PR| = 6<br>
30
(c) Claire is at a point c on the top of a cliff. The point B is at the base of the cliff. The height of the cliff is 35 m, as shown in the diagram. She wishes to find |BA|, the distance from the
base of the cliff to the base of the lighthouse. She measured
∠ DCA and found it to be 41º. CD is parallel to BA.
(i) Find |∠ BAC|. 35 m 41º D B C A Alternate angles |∠ BAC| = 41º 41º<br>
base of the cliff to the base of the lighthouse. She measured
∠ DCA and found it to be 41º. CD is parallel to BA.
(i) Find |∠ BAC|. 35 m 41º D B C A Alternate angles |∠ BAC| = 41º 41º<br>
31
40 m (c) Claire is at a point c on the top of a cliff. The point B is at the base of the cliff. The height of the cliff is 35 m, as shown in the diagram. She wishes to find |BA|, the distance from the
base of the cliff to the base of the lighthouse. She measured
∠ DCA and found it to be 41º. CD is parallel to BA.
(ii) Find, to the nearest metre, |BA|, the distance from the base of
the cliff to the base of the lighthouse. tan 41° = 35 m 41º D OPP |BA| 35 B C A 41º ADJ tan 41° = = 40·262… = 40 metres<br>
base of the cliff to the base of the lighthouse. She measured
∠ DCA and found it to be 41º. CD is parallel to BA.
(ii) Find, to the nearest metre, |BA|, the distance from the base of
the cliff to the base of the lighthouse. tan 41° = 35 m 41º D OPP |BA| 35 B C A 41º ADJ tan 41° = = 40·262… = 40 metres<br>
32
Question 7 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
33
(a) The right-angled triangle ABC has measurements as shown. HYP ADJ C A B A 17 15 8 (i) Write down the length of the side adjacent to angle A. 15 (ii) Write down the value of cos A, as a fraction. 17 Length of the side adjacent to angle A = 15<br>
34
–––––––––– 37 º (ii) Using your calculator, write
down the value of sin 37º
correct to one decimal place. (iii) Hence find x, the value of | PR |. hypotenuse (b) In the right-angled triangle PQR,
| PQ | = 12, |∠ PQR | = 37º. Let x = | PR |. P Q R (i) Using the diagram, write down the
value of sin 37° as a fraction. sin 37º = 0·6018…. 12 HYP OPP opposite sin 37º = x 12 0·6 = × x = 7·2 x<br>
down the value of sin 37º
correct to one decimal place. (iii) Hence find x, the value of | PR |. hypotenuse (b) In the right-angled triangle PQR,
| PQ | = 12, |∠ PQR | = 37º. Let x = | PR |. P Q R (i) Using the diagram, write down the
value of sin 37° as a fraction. sin 37º = 0·6018…. 12 HYP OPP opposite sin 37º = x 12 0·6 = × x = 7·2 x<br>
35
(c) Ciara wished to measure the width of a river.
She was at A on the riverbank, directly opposite B on the other
bank. Ciara walked from A to C, along the riverbank, at an
average speed of 1·5 m/s. It took Ciara 30 seconds to reach C.
She then measured ∠ ACB and found it to be 25º . (i) Calculate | AC| , the distance walked by Ciara. Distance = Speed × Time = 1·5 × 30 = 45 metres 45 m<br>
She was at A on the riverbank, directly opposite B on the other
bank. Ciara walked from A to C, along the riverbank, at an
average speed of 1·5 m/s. It took Ciara 30 seconds to reach C.
She then measured ∠ ACB and found it to be 25º . (i) Calculate | AC| , the distance walked by Ciara. Distance = Speed × Time = 1·5 × 30 = 45 metres 45 m<br>
36
45 (ii) Hence, calculate | AB| , the width of the river.
Give your answer correct to the nearest metre. 45 m OPP ADJ tan 25° = | AB| tan 25° = | AB| = 20·983… 1 m (c) Ciara wished to measure the width of a river.
She was at A on the riverbank, directly opposite B on the other
bank. Ciara walked from A to C, along the riverbank, at an
average speed of 1·5 m/s. It took Ciara 30 seconds to reach C.
She then measured ∠ ACB and found it to be 25º .<br>
Give your answer correct to the nearest metre. 45 m OPP ADJ tan 25° = | AB| tan 25° = | AB| = 20·983… 1 m (c) Ciara wished to measure the width of a river.
She was at A on the riverbank, directly opposite B on the other
bank. Ciara walked from A to C, along the riverbank, at an
average speed of 1·5 m/s. It took Ciara 30 seconds to reach C.
She then measured ∠ ACB and found it to be 25º .<br>
37
Question 8 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
38
(a) The right-angled triangle ABC has measurements as shown. HYP OPP C A A B 13 12 5 (i) Write down the length of the side opposite the angle A. 5 (ii) Write down the value of sin A, as a fraction. 5 13<br>
39
–––––––––– (ii) Hence find the measure of ∠ PQR,
correct to the nearest degree. hypotenuse (b) In the right-angled triangle PQR,
| PQ | = 10 and | PR | = 4. P Q R (i) Find the value of cos ∠ PQR . ∠ PQR = cos–1 0·4 4 HYP ADJ adjacent cos ∠ PQR = 4 10 = 66·421... 10 = 0·4 = 66º<br>
correct to the nearest degree. hypotenuse (b) In the right-angled triangle PQR,
| PQ | = 10 and | PR | = 4. P Q R (i) Find the value of cos ∠ PQR . ∠ PQR = cos–1 0·4 4 HYP ADJ adjacent cos ∠ PQR = 4 10 = 66·421... 10 = 0·4 = 66º<br>
40
(c) An aeroplane leaves the ground at an angle of 20°
to the runway. Its speed is 28 m/sec. (i) How far does the aeroplane travel
in the first 30 seconds? Take-off point 20º Distance = Speed × Time = 28 × 30 = 840 metres 840 m<br>
to the runway. Its speed is 28 m/sec. (i) How far does the aeroplane travel
in the first 30 seconds? Take-off point 20º Distance = Speed × Time = 28 × 30 = 840 metres 840 m<br>
41
(c) An aeroplane leaves the ground at an angle of 20°
to the runway. Its speed is 28 m/sec. h (ii) What is its height above the ground after the
first 30 seconds?
Write your answer to
the nearest metre. HYP 840 m 20º OPP hypotenuse opposite sin 20º = h 840 h = 287·296.. metres<br>
to the runway. Its speed is 28 m/sec. h (ii) What is its height above the ground after the
first 30 seconds?
Write your answer to
the nearest metre. HYP 840 m 20º OPP hypotenuse opposite sin 20º = h 840 h = 287·296.. metres<br>
42
Question 9 Junior Certificate
Ordinary Level<br>
Ordinary Level<br>
43
5 (a) The triangle abc has measurements as shown. HYP ADJ b A c a 5 4 3 (i) Write down the value of cos A. (ii) Write down the value of tan A. 4 OPP 4 3<br>
44
(b) A vertical building is 8 m high.
It casts a shadow three times its height on horizontal ground. B 8 m (i) Write down the length of the shadow. 3 × 8 = 24 m (ii) Find B, the angle of elevation of the sun, correct
to the nearest degree. 24 m –––––––– adjacent opposite tan B = 8 24 ADJ OPP B = tan – 1 = 18 ·4349… °<br>
It casts a shadow three times its height on horizontal ground. B 8 m (i) Write down the length of the shadow. 3 × 8 = 24 m (ii) Find B, the angle of elevation of the sun, correct
to the nearest degree. 24 m –––––––– adjacent opposite tan B = 8 24 ADJ OPP B = tan – 1 = 18 ·4349… °<br>
45
(c) A vertical flagpole [PQ], 12 m high, is supported by a
cable [QR] as shown in the diagram. P Q R 12 m 30° (i) Given that | ∠ QRP | = 30°, find the length of the cable [QR]. x HYP OPP hypotenuse opposite sin 30º 12 x 0·5 –––––––––– = = 24 m 24 m<br>
cable [QR] as shown in the diagram. P Q R 12 m 30° (i) Given that | ∠ QRP | = 30°, find the length of the cable [QR]. x HYP OPP hypotenuse opposite sin 30º 12 x 0·5 –––––––––– = = 24 m 24 m<br>
46
576 – 144 432 24 m (c) A vertical flagpole [PQ], 12 m high, is supported by a
cable [QR] as shown in the diagram. P Q R 12 m 30° HYP OPP (ii) How far is R from P, the foot of the flagpole?
Give your answer correct to one decimal place. Pythagoras’ Theorem a2 + b2 = c2 y2 + 122 = 242 y2 + 144 = 576 y 2 = y = 20· 784… 8 m y<br>
cable [QR] as shown in the diagram. P Q R 12 m 30° HYP OPP (ii) How far is R from P, the foot of the flagpole?
Give your answer correct to one decimal place. Pythagoras’ Theorem a2 + b2 = c2 y2 + 122 = 242 y2 + 144 = 576 y 2 = y = 20· 784… 8 m y<br>