1. hypotenuse opposite adjacent Remember
This is opposite the angle.
This is opposite the right-angle
There are three ratios that you need to learn:
hypotenuse
opposite
adjacent
This is next to the angle
Remember
Where are the hypotenuse, adjacent and opposite lengths.

2. Finding a length hypotenuse opposite adjacent
You need to label all the lengths of the triangle.
hypotenuse
10cm
The adjacent is not needed in this
question; the question is only using
opposite and hypotenuse.
opposite
adjacent

3. Finding a length hypotenuse opposite
You need to label all the lengths of the triangle.
hypotenuse
10cm
The adjacent is not needed in this
question; the question is only using
opposite and hypotenuse.
opposite

4. Finding a length hypotenuse opposite 10cm
We need to use the sine ratio and substitute the values/variables that we know.
hypotenuse
10cm
opposite

5. Finding a length hypotenuse opposite 10cm OR
We need to solve the equation (*).
hypotenuse
10cm
opposite
We can think of this as: OR
rearrange the equation (*) by multiplying both sides by x.

6. These two triangles are very different.
The variable x will be on the denominator
This is an example of where we have to solve an equation with the variable on the denominator, so multiply both sides by x.
still need to get x on its own, so divide both sides by sin30

7. 2. Pick the correct formula
1. label the triangle
opposite
15cm
3. Substitute in the correct values
adjacent
4. Rearrange to find x

8. 15cm
opposite
adjacent

10. The Trigonometric RatiosB C
hypotenuse
opposite
adjacent
Make up a Mnemonic! S O C H A T
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11. The Trigonometric Ratios (Finding an unknown side).
Use your worksheet and a ruler to measure as accurately as possible the sides of each triangle above. Find approximate values for the sine, cosine and tangent ratios for an angle of 30o.
True Values (2 dp)
The fixed values of these ratios were calculated for every angle and stored in a table of sines, cosines and tangents. Nowadays a calculator computes each value instead. Because these values are constant we can use them to find unknown sides.
To Trig Tables >
Sin 30o = 0.50
Cos 30o = 0.87
Tan 30o = 0.58
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12. The Trigonometric Ratios (Finding an unknown side).
Sin 30o = 0.50
Cos 30o = 0.87
Tan 30o = 0.58
True Values (2 dp)
The Trigonometric Ratios (Finding an unknown side).
30o
For example, anytime we come across a right-angled triangle containing an angle of 30o we can find an unknown side if we are given the value of one other. h
30o
75 m
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13. The Trigonometric Ratios (Finding an unknown side).
Example 1. In triangle ABC find side CB.
70o A B C
12 cm
Diagrams not to scale. S O H C A T
Opp
Example 2. In triangle PQR find side PQ.
22o P Q R
7.2 cm S O H C A T
Example 3. In triangle LMN find side MN.
75o L M N
4.3 m S O H C A T
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14. The Trigonometric Ratios (Finding an unknown angle).
Anytime we come across a right-angled triangle containing 2 given sides we can calculate the ratio of the sides then look up (or calculate) the angle that corresponds to this ratio.
Sin 30o = 0.50
Cos 30o = 0.87
Tan 30o = 0.58
True Values (2 dp) xo
43.5 m
75 m S O H C A T
30o
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15. The Trigonometric Ratios (Finding an unknown angle).
Example 1. In triangle ABC find angle A. A B C
12 cm
11.3 cm
Diagrams not to scale. S O H C A T
Sin-1(11.3  12) =
Key Sequence S O H C A T
Example 2. In triangle LMN find angle N. L M N
4.3 m
1.2 m
Tan-1(4.3  1.2) =
Key Sequence
Example 3. In triangle PQR find angle Q. P Q R
7.2 cm
7.8 cm S O H C A T
Cos-1(7.2  7.8) =
Key Sequence
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16. Applications of Trigonometry
A boat sails due East from a Harbour (H), to a marker buoy (B), 15 miles away. At B the boat turns due South and sails for 6.4 miles to a Lighthouse (L). It then returns to harbour. Make a sketch of the trip and calculate the bearing of the harbour from the lighthouse to the nearest degree. H B L
15 miles
6.4 miles
SOH CAH TOA
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17. Applications of Trigonometry
A 12 ft ladder rests against the side of a house. The top of the ladder is 9.5 ft from the floor. Calculate the angle that the foot of ladder makes with the ground.
Applications of Trigonometry Lo
SOH CAH TOA
16/06/2018 17

18. Applications of Trigonometry
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 430 miles North to a point P before turning left and flying for 570 miles to a second point Q, West of W. It then returns to base.
(a) Make a sketch of the flight.
(b) Find the bearing of Q from P.
Applications of Trigonometry
Not to Scale P
570 miles W
430 miles Q
16/06/2018
SOH CAH TOA 18

19. Angles of Elevation and Depression.
An angle of elevation is the angle measured upwards from a horizontal to a fixed point. The angle of depression is the angle measured downwards from a horizontal to a fixed point.
Horizontal
Horizontal
25o
Angle of depression
Explain why the angles of elevation and depression are always equal.
25o
Angle of elevation
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20. Applications of Trigonometry
A man stands at a point P, 45 m from the base of a building that is 20 m high. Find the angle of elevation of the top of the building from the man.
Applications of Trigonometry
20 m
45 m P
SOH CAH TOA
16/06/2018 20

21. A 25 m tall lighthouse sits on a cliff top, 30 m above sea level
A 25 m tall lighthouse sits on a cliff top, 30 m above sea level. A fishing boat is seen 100m from the base of the cliff, (vertically below the lighthouse). Find the angle of depression from the top of the lighthouse to the boat.
100 m
55 m D C D
Or more directly since the angles of elevation and depression are equal.
SOH CAH TOA
16/06/2018 21

22. A 22 m tall lighthouse sits on a cliff top, 35 m above sea level
A 22 m tall lighthouse sits on a cliff top, 35 m above sea level. The angle of depression of a fishing boat is measured from the top of the lighthouse as 30o. How far is the fishing boat from the base of the cliff?
x m
57 m
30o
60o
Or more directly since the angles of elevation and depression are equal.
30o
SOH CAH TOA
16/06/2018 22

23. The origins of trigonometry are closely tied up with problems involving circles. One particular problem is that of finding the lengths of chords subtended by different angles at the centre of a circle. The Arabs called the half chord (“ardha-jya”). This became mis-interpreted and mis-translated over the centuries and eventually ended up as “sinus” in Latin, meaning cove or bay. Other derivations include: bulge, bosom, sinus, cavity, nose and skull. The cosinus simply means the compliment of the sinus, since SinA = Cos (90 – A) (Sin 60 = Cos 30, Sin 70 = Cos 20 etc)
Chord M
Angular Bisector
Radius P O Q
Half Chord PM (Sinus)
OM (Cosinus)
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24. The following diagrams show the relationships between the 3 trigonometric ratios for a circle of radius 1 unit. Tangent means “To touch”
Radius P O Q
Half Chord PM
Sin  = O/H = PM/1 = PM
Chord
Cos  = A/H = OM/1 = OM M
Angular Bisector o
Tan  = PT/1 = PT
Tangent T o P O 1 M P M O 1  P O T  1
Sinus
Cosinus M O P 1 
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25. The Trigonometric Ratios (Finding an unknown side).
Use your worksheet and a ruler to measure as accurately as possible the sides of each triangle above. Find approximate values for the sin, cos and tan ratios for an angle of 30o.
Sin 30o =
Cos 30o =
Tan 30o =
Estimated Values
Worksheet 1
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26. Trig Tables
Click to go back
16/06/2018 26

27. The Trigonometric Ratios (Finding an unknown side).
Example 1. In triangle ABC find side CB.
70o A B C
12 cm
Opp
Example 2. In triangle PQR find side PQ.
22o P Q R
7.2 cm
Example 3. In triangle LMN find side MN.
75o L M N
4.3 m
Worksheet 2
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28. The Trigonometric Ratios (Finding an unknown angle).
Example 1. In triangle ABC find angle A. A B C
12 cm
11.3 cm
The Trigonometric Ratios (Finding an unknown angle).
Example 2. In triangle LMN find angle N. L M N
4.3 m
Example 3. In triangle PQR find angle Q. P Q R
7.2 cm
7.8 cm
1.2 m
Worksheet 3
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29. Do this at the start of all Trig problems
Label each side of the following triangles. 2) 1) 3)
Hyp
Opp
Opp
Opp
Hyp
Adj
Adj
Hyp
Adj 5) 4)
Adj 6)
Opp
Opp
Adj
Hyp
Opp
Hyp
Hyp
Adj 7)
Hyp
Adj
Opp
Do this at the start of all Trig problems 29

30. 90°
80°
70°
60°
50°
40°
30°
20°
10° 0° 30

31. Sin 300 = 0.5
2.3
4.6
= 0.5
= 0.5 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 18 17 16
HYP
OPP
4.6
9.2
ADJ
6.9
= 0.5
8.7
= 0.5
13.9 15
17.3
8.7
17.3 8 9 7
1cm 2 3 4 5 6
Sin angle = OPP
HYP 12
13.9
6.9 8
9.2
4.6
4.6
30º
2.3 4 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 31

32. Cos 300 = 0.866 4 8
= 0.87
= 0.87
OPP 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 18 17 16
HYP
4.6
9.2
ADJ 12
= 0.86 15
= 0.87
13.9 15
17.3
8.7
17.3 8 9 7
1cm 2 3 4 5 6
Cos angle = ADJ
HYP 12
13.9
6.9 8
9.2
4.6
4.6
30º
2.3 4 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 32

33. Tan 300 = 0.577
2.3
4.6
= 0.575
= 0.575
OPP 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 18 17 16 4 8
HYP
ADJ
6.9
= 0.575
8.7
= 0.58 12 15
17.3
8.7 15 8 9 7
1cm 2 3 4 5 6
Tan angle = OPP
ADJ 12
13.9
6.9 8
9.2
4.6
4.6
30º
2.3 4 15 14 13 12 11 10 8 9 7
1cm 2 3 4 5 6 33

34. You need to memorise this !
Some Old Hags Can’t Always Hide Their Old Age.
Sin angle = Opp
Hyp
Cos angle = Adj
Hyp
Tan angle = Opp
Adj
Opp
Opp
Hyp
Hyp
Adj
Adj
You need to memorise this ! 34
Menu

35. Working Out x tan angle = Opp Adj 8cm 4.62 0.5774 Tan 300 = x
Working out a Length ( Unknown on Top )
Some Old Hags Can’t Always Hide Their Old Age. x
Hyp
tan angle = Opp
Opp
300
Adj
Adj
8cm
4.62
0.5774
Tan 300 = x
tan 300 = x x 8 8
tan 300 = x
4.62 = x
= x
Working Out 8 8
8 x = x
tan 300 = x
4.62 = x 8
8 tan 30 0 = x
4.62 = x 35
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36. Calculate the unknown lengths x.
(Give your answers to 1 decimal place.) 1) 2) 3) 4) 15 7 10 x x x
40°
50°
45°
30° 6
5.0
11.5 x
7.1
3.5 5) 6) 7) 8) x x x
35°
50°
75° x 7 40 50 8
25°
3.3
32.8
12.9
6.1 9)
10)
11)
12)
50° 14 x 30
40°
70°
55° x 25 x x
10.7 15
5.1
21.0
24.6 36
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Answers

37. Working Out Working Out A Length ( Unknown on Bottom ) x
Some Old Hags Can’t Always Hide Their Old Age. x
Cos angle = Adj
Opp
Hyp
Hyp
400
Cos 400
0.766
7.83 6
Adj =
Cos 400 = 6 x x
6 cm
Cos 400
= 7.83
= 6
0.766 = 6 x
Working Out x
x =
0.766
Cos = 6 x
x = 7.83
x =
Cos 400 37
x = 7.83
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38. Calculate the unknown lengths x.
(Give your answers to 1 decimal place.) 1) 2) 3) 4) x x x 12 8 8
40°
50°
60°
45° x
9.5
15.7 10
20.0
11.3 5) 6) 7) 8) x x
30°
20°
55° 15 5 2 7 x x
40°
26.0
2.4
9.1
13.7 9)
10)
11)
12)
40° x x x 3
35°
60°
40° 12 x 20 5
18.7
1.7
28.6
6.5 38
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Answers

39. Working Out 8 ÷ 11 Working Out An Angle. 11 cm Cos angle = Adj Opp Hyp
Some Old Hags Can’t Always Hide Their Old Age.
11 cm
Cos angle = Adj
Opp
Hyp
Hyp
Cos x = 8
Cos x =
x =
Cos-1
43.340
0.7273 8 x0
Adj 11 11
8 cm
Working Out
Cos x =
8 ÷ 11
x = Cos
X =
Always inverse
sin, cos or tan at the end. 39
Menu

40. Calculate the unknown angles x.
(Give your answers to 1 decimal place.) 1) 2) 3) 4) 10 15 10 7 8 x° x° x° x°
53.1° 6 8
36.9°
66.4° 3
66.8° 5) 6) 7) 8) 17 8 x° x° x° 12 8 1 30 11 x° 10
55.5°
46.7°
84.3°
56.3° 9)
10)
11)
12) x° 20 24 14 x° x° x° 16 4 7 10
41.8° 30
27.8°
60.3°
30.0° 40
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Answers

41. Don’t forget to put calculators in DEGREES mode !
Trigonometry Summary.
Some Old Hags Can’t Always Hide Their Old Age.
Don’t forget to put calculators in
DEGREES mode !
Unknown on Top
Unknown on Bottom
Calculating an Angle
9 cm x
7 cm
Hyp
9 cm
Opp
Opp
Opp
Hyp
Hyp
Adj
500
Adj
400
Adj x0 x
7 cm
Cos 500 = x
Sin 400 = 7
Tan x0 = 9 9 x 7
= x
= 7
Tan x =
9 cos = x
x = 7 9 x
Sin 400
x = Tan
9 x = x
5.79 = x
x =
x = 10.89
0.6429
x =
5.79 = x
x = 41
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42. Calculate the unknown lengths or angles x.
(Give your answers to 1 decimal place.) 1) 2) 3) 4) x x 20 10 x 6
60°
50° x°
70° 4
6.9
13.1
17.5° 5
14.6 5) 6) 7) 8) 12 3 x
40°
30° x° x 5 9 15 x
40°
7.6
9.6
59°
13.9 9)
10)
11)
12) 40 x°
200 x 20 8
60°
45° x° x 50 5
78.5°
8.0
28.9
14.5° 42
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Answers

43. A = Cos o x H
Cosine Rule
To find an adjacent side we need 1 side (hypotenuse) and the included angle.
9 cm
12 cm
60°
75° a a
A = Cos ° x H
A = Cos 75° x 12
A = x 12
A = 3.1 cm
A = Cos ° x H
A = Cos 60° x 9
A = 0.5 x 9
A = 4.5 cm

44. Cos o = A ÷ H
Cosine Rule
To find an unknown angle we need 2 sides the adjacent and the hypotenuse
5 cm
15 cm x° y°
2.5 cm
12cm
Cos ° = A ÷ H
Cos ° = 12 ÷ 15
Cos ° = 0.8
Shift (Inv) (2ndF) Cos = 36.9°
Cos ° = O ÷ H
Cos° = 2.5 ÷ 5
Cos° = 0.5
Shift (Inv) (2ndF) Cos = 60°

45. O = Sin o x H
Sine Rule
To find an opposite side we need 1 side (hypotenuse) and the included angle.
5 cm
10 cm o o
30°
75°
O = Sin ° x H
O = Sin 30° x 5
O = 0.5 x 5
O = 2.5 cm
O = Sin ° x H
O = Sin 75° x 10
O = x 10
O = 9.7cm

46. Sin o = O ÷ H
Sine Rule
To find an unknown angle we need 2 sides the opposite and the hypotenuse
5 cm
15 cm
2.5 cm
12cm x° y°
Sin ° = A ÷ H
Sin ° = 12 ÷ 15
Sin ° = 0.8
Shift (Inv) (2ndF) Sin = 53.1°
Sin ° = O ÷ H
Sin ° = 2.5 ÷ 5
Sin ° = 0.5
Shift (Inv) (2ndF) Sin = 30°

47. To find an unknown angle we need 2 sides the opposite and the adjacent
Tan o = O ÷ A
Tangent Rule
To find an unknown angle we need 2 sides the opposite and the adjacent
2.5 cm
12cm x° y°
5 cm
15 cm
Tan ° = A ÷ H
Tan ° = 12 ÷ 15
Tan ° = 0.8
Shift (Inv) (2ndF) Tan = 38.6°
Tan ° = O ÷ H
Sin ° = 2.5 ÷ 5
Sin ° = 0.5
Shift (Inv) (2ndF) Tan = 26.6°

48. O = Tan o x A
Tangent Rule
To find an opposite side we need 1 side (adjacent) and the included angle. a a
45°
75°
9 cm
6 cm
O = Tan ° x A
O = Tan 45° x 9
O = 1 x 9
O = 9 cm
A = Tan ° x H
A = Tan 75° x 6
A = x 6
A = 22.4 cm

49. (1) Sin, Cos or Tan? Answer: Tan x 35o 7 S H O C H A T A O
You know the adjacent and want the opposite. x
35o 7
S H O
C H A
T A O

50. (2) Sin, Cos or Tan? Answer: Sin x 10 40o S H O C H A T A O
You know the opposite and want the hypotenuse. x 10
40o
S H O
C H A
T A O

51. (3) Sin, Cos or Tan? Answer: Cos 35o 20 S H O C H A T A O
You know the adjacent and want the hypotenuse.
35o x 20
S H O
C H A
T A O

52. (4) Sin, Cos or Tan? Answer: Sin x 12 38o S H O C H A T A O
You know the hypotenuse and want the opposite. x 12
38o
S H O
C H A
T A O

53. (5) Sin, Cos or Tan? Answer: Sin 18 10 o S H O C H A T A O
You know the opposite and the hypotenuse. You want to find the angle. 18 10 o
S H O
C H A
T A O

54. (6) Sin, Cos or Tan? Answer: Sin 15 10 o S H O C H A T A O
You know the opposite and the hypotenuse. And want to know the angle 15 10 o
S H O
C H A
T A O

55. (7) Sin, Cos or Tan? Answer: Cos 21o x S H O C H A T A O
You know the hypotenuse and want the adjacent.
21o
100 x
S H O
C H A
T A O

56. (8) Sin, Cos or Tan? Answer: Tan 24 37o x S H O C H A T A O
You know the opposite and want the adjacent. 24
37o x
S H O
C H A
T A O

57. (9) Sin, Cos or Tan? Answer: Tan 20 42o x S H O C H A T A O
You know the opposite and want the adjacent. 20
42o x
S H O
C H A
T A O

58. (10) Sin, Cos or Tan? Answer: Sin 400 200 o S H O C H A T A O
You know the opposite and the hypotenuse. You want to find the angle.
400
200 o
S H O
C H A
T A O

59. (1) Trig Solutions Tan x 35o 7 S H O C H A T A O
You know the adjacent and want the opposite.
Opp = Tan x Adj
x = tan 35 x 7
x = 4.9 (1.d.p.) x
35o 7
S H O
C H A
T A O

60. (2) Trig Solutions Sin x 10 40o S H O C H A T A O
You know the opposite and want the hypotenuse. x
Hyp = Opp/sin40
x = 15.6 cm 10
40o
S H O
C H A
T A O

61. (3) Trig Solutions Cos 35o 20 S H O C H A T A O
You know the adjacent and want the hypotenuse.
35o x
Hyp = Adj/cos 35
x = 24.4 20
S H O
C H A
T A O

62. (4) Trig Solutions Sin x 12 38o S H O C H A T A O
You know the hypotenuse and want the opposite. x 12
38o
Opp = Hyp x sin 38
x = 12 x sin 38
x = 24.4
S H O
C H A
T A O

63. (5) Trig Solutions Tan 20 42o x S H O C H A T A O
You know the opposite and want the adjacent. 20
Adj = Opp/tan 42
x = 22.2
42o x
S H O
C H A
T A O

64. (6) Trig Solutions Sin 15 10 o S H O C H A T A O
You know the opposite and the hypotenuse. And want to know the angle 15 10 o
S H O
C H A
T A O

65. (7) Trig Solutions Cos 21o x S H O C H A T A O
You know the hypotenuse and want the adjacent.
21o
100
Adj = Hyp x Cos 21
x = 100 x cos 21
x = 93.4 x
S H O
C H A
T A O

66. (8) Trig Solutions Tan 24 37o x S H O C H A T A O
You know the opposite and want the adjacent. 24
Adj = Opp/tan 37
x = 31.8
37o x
S H O
C H A
T A O

67. (9) Trig Solutions Sin 18 10 o S H O C H A T A O
You know the opposite and the hypotenuse. You want to find the angle. 18 10 o
S H O
C H A
T A O

68. (10) Trig Solutions Sin 400 200 o S H O C H A T A O
You know the opposite and the hypotenuse. You want to find the angle.
400
200 o
S H O
C H A
T A O

69. Applications of Trigonometry
A boat sails due East from a Harbour (H), to a marker buoy (B), 15 miles away. At B the boat turns due South and sails for 6.4 miles to a Lighthouse (L). It then returns to harbour. Make a sketch of the trip and calculate the bearing of the harbour from the lighthouse to the nearest degree. H B L
15 miles
6.4 miles
SOH CAH TOA

70. Applications of Trigonometry
A 12 ft ladder rests against the side of a house. The top of the ladder is 9.5 ft from the floor. Calculate the angle that the foot of ladder makes with the ground.
Applications of Trigonometry Lo
SOH CAH TOA

71. Applications of Trigonometry
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 430 miles North to a point P before turning left and flying for 570 miles to a second point Q, West of W. It then returns to base.
(a) Make a sketch of the flight.
(b) Find the bearing of Q from P.
Applications of Trigonometry
Not to Scale P
570 miles W
430 miles Q
SOH CAH TOA

72. Angles of Elevation and Depression.
An angle of elevation is the angle measured upwards from a horizontal to a fixed point. The angle of depression is the angle measured downwards from a horizontal to a fixed point.
Horizontal
Horizontal
25o
Angle of depression
Explain why the angles of elevation and depression are always equal.
25o
Angle of elevation

73. Applications of Trigonometry
A man stands at a point P, 45 m from the base of a building that is 20 m high. Find the angle of elevation of the top of the building from the man.
Applications of Trigonometry
20 m
45 m P
SOH CAH TOA

74. A 25 m tall lighthouse sits on a cliff top, 30 m above sea level
A 25 m tall lighthouse sits on a cliff top, 30 m above sea level. A fishing boat is seen 100m from the base of the cliff, (vertically below the lighthouse). Find the angle of depression from the top of the lighthouse to the boat.
100 m
55 m D C D
Or more directly since the angles of elevation and depression are equal.
SOH CAH TOA

75. A 22 m tall lighthouse sits on a cliff top, 35 m above sea level
A 22 m tall lighthouse sits on a cliff top, 35 m above sea level. The angle of depression of a fishing boat is measured from the top of the lighthouse as 30o. How far is the fishing boat from the base of the cliff?
x m
57 m
30o
60o
Or more directly since the angles of elevation and depression are equal.
30o
SOH CAH TOA