How to Use This Page
1. Read the Key Concepts recap to refresh the formulas and standard values you need. 2. Attempt each question on your own first. Write your working in a notebook. 3. Compare your steps with the full worked solution shown here. 4. Mark where you went wrong and redo that question. 5. Use the Exam Checklist before your test to verify you have covered every type.
Key Concepts (Quick Recap)
Six Trigonometric Ratios
For a right triangle with angle theta:
| Ratio | Formula |
|---|---|
| sin theta | Opposite / Hypotenuse |
| cos theta | Adjacent / Hypotenuse |
| tan theta | Opposite / Adjacent |
| cosec theta | Hypotenuse / Opposite |
| sec theta | Hypotenuse / Adjacent |
| cot theta | Adjacent / Opposite |
Standard Angle Values
| theta | 0 degrees | 30 degrees | 45 degrees | 60 degrees | 90 degrees |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/sqrt(2) | sqrt(3)/2 | 1 |
| cos | 1 | sqrt(3)/2 | 1/sqrt(2) | 1/2 | 0 |
| tan | 0 | 1/sqrt(3) | 1 | sqrt(3) | undefined |
| cosec | undefined | 2 | sqrt(2) | 2/sqrt(3) | 1 |
| sec | 1 | 2/sqrt(3) | sqrt(2) | 2 | undefined |
| cot | undefined | sqrt(3) | 1 | 1/sqrt(3) | 0 |
Fundamental Identities
- sin squared theta + cos squared theta = 1
- 1 + tan squared theta = sec squared theta
- 1 + cot squared theta = cosec squared theta
Complementary Angle Relations
- sin(90 degrees minus theta) = cos theta
- cos(90 degrees minus theta) = sin theta
- tan(90 degrees minus theta) = cot theta
Heights and Distances
- Angle of elevation: angle measured upward from horizontal to the line of sight.
- Angle of depression: angle measured downward from horizontal to the line of sight.
- Always draw a labelled diagram before solving.
Exercise 6.1 — Trigonometric Ratios
Q1. If sin A = 3/5, find all other trigonometric ratios of angle A.
Given: sin A = 3/5 = Opposite / Hypotenuse
Let Opposite = 3, Hypotenuse = 5
By Pythagoras theorem:
Adjacent = sqrt(Hypotenuse squared - Opposite squared)
= sqrt(25 - 9)
= sqrt(16)
= 4
Now write all ratios:
cos A = Adjacent / Hypotenuse = 4/5
tan A = Opposite / Adjacent = 3/4
cosec A = Hypotenuse / Opposite = 5/3
sec A = Hypotenuse / Adjacent = 5/4
cot A = Adjacent / Opposite = 4/3
Q2. If cos theta = 12/13, find sin theta, tan theta, and sec theta.
Given: cos theta = 12/13 = Adjacent / Hypotenuse
Let Adjacent = 12, Hypotenuse = 13
By Pythagoras theorem:
Opposite = sqrt(Hypotenuse squared - Adjacent squared)
= sqrt(169 - 144)
= sqrt(25)
= 5
Now:
sin theta = Opposite / Hypotenuse = 5/13
tan theta = Opposite / Adjacent = 5/12
sec theta = Hypotenuse / Adjacent = 13/12
Q3. If tan A = 1/sqrt(3), find the value of sin A cos A + cos A sin A (i.e. 2 sin A cos A).
Given: tan A = 1/sqrt(3) = Opposite / Adjacent
Let Opposite = 1, Adjacent = sqrt(3)
Hypotenuse = sqrt(Opposite squared + Adjacent squared)
= sqrt(1 + 3)
= sqrt(4)
= 2
So:
sin A = Opposite / Hypotenuse = 1/2
cos A = Adjacent / Hypotenuse = sqrt(3)/2
Now:
2 sin A cos A = 2 x (1/2) x (sqrt(3)/2)
= 2 x sqrt(3)/4
= sqrt(3)/2
Q4. If 3 cot A = 4, check whether (1 minus tan squared A) / (1 + tan squared A) = cos squared A minus sin squared A.
Given: 3 cot A = 4 => cot A = 4/3
Since cot A = Adjacent / Opposite,
Let Adjacent = 4, Opposite = 3
Hypotenuse = sqrt(9 + 16) = sqrt(25) = 5
sin A = 3/5, cos A = 4/5, tan A = 3/4
Left side:
(1 - tan squared A) / (1 + tan squared A)
= (1 - 9/16) / (1 + 9/16)
= (7/16) / (25/16)
= 7/25
Right side:
cos squared A - sin squared A
= 16/25 - 9/25
= 7/25
Both sides are equal.
Q5. Evaluate: sin 60 degrees cos 30 degrees + sin 30 degrees cos 60 degrees.
sin 60 degrees = sqrt(3)/2, cos 30 degrees = sqrt(3)/2
sin 30 degrees = 1/2, cos 60 degrees = 1/2
= sin 60 degrees x cos 30 degrees + sin 30 degrees x cos 60 degrees
= (sqrt(3)/2)(sqrt(3)/2) + (1/2)(1/2)
= 3/4 + 1/4
= 4/4
= 1
(Also equals sin(60 + 30) = sin 90 degrees = 1)
Q6. If sin theta + cos theta = sqrt(3), then prove that tan theta + cot theta = 1.
Given: sin theta + cos theta = sqrt(3)
Squaring both sides:
sin squared theta + cos squared theta + 2 sin theta cos theta = 3
Since sin squared theta + cos squared theta = 1:
1 + 2 sin theta cos theta = 3
2 sin theta cos theta = 2
sin theta cos theta = 1
Now:
tan theta + cot theta = (sin theta / cos theta) + (cos theta / sin theta)
= (sin squared theta + cos squared theta) / (sin theta cos theta)
= 1 / (sin theta cos theta)
= 1 / 1
= 1
Q7. In a right triangle ABC, right angled at B, if tan A = 1, then verify that sin A cos A = 1/2.
Given: tan A = 1 = Opposite / Adjacent
Let Opposite = 1, Adjacent = 1 (since ratio is 1:1)
Hypotenuse = sqrt(1 + 1) = sqrt(2)
sin A = 1/sqrt(2), cos A = 1/sqrt(2)
sin A cos A = (1/sqrt(2)) x (1/sqrt(2))
= 1/2
Exercise 6.2 — Trigonometric Identities
Q1. Prove that (1 minus cos squared A) cosec squared A = 1.
LHS = (1 - cos squared A) x cosec squared A
= sin squared A x cosec squared A [since 1 - cos squared A = sin squared A]
= sin squared A x (1/sin squared A) [since cosec A = 1/sin A]
= 1
= RHS
Q2. Prove that (1 + tan squared A) cos squared A = 1.
LHS = (1 + tan squared A) x cos squared A
= sec squared A x cos squared A [since 1 + tan squared A = sec squared A]
= (1/cos squared A) x cos squared A [since sec A = 1/cos A]
= 1
= RHS
Q3. Prove that tan A + cot A = sec A cosec A.
LHS = tan A + cot A
= sin A / cos A + cos A / sin A
= (sin squared A + cos squared A) / (sin A cos A)
= 1 / (sin A cos A) [since sin squared A + cos squared A = 1]
= (1/cos A) x (1/sin A)
= sec A x cosec A
= RHS
Q4. Prove that (sin A minus cos A) squared = 1 minus 2 sin A cos A.
LHS = (sin A - cos A) squared
= sin squared A - 2 sin A cos A + cos squared A
= (sin squared A + cos squared A) - 2 sin A cos A
= 1 - 2 sin A cos A [since sin squared A + cos squared A = 1]
= RHS
Q5. Prove that sqrt((1 minus cos A) / (1 + cos A)) = cosec A minus cot A.
LHS = sqrt((1 - cos A) / (1 + cos A))
Multiply numerator and denominator by (1 - cos A):
= sqrt((1 - cos A) squared / ((1 + cos A)(1 - cos A)))
= sqrt((1 - cos A) squared / (1 - cos squared A))
= sqrt((1 - cos A) squared / sin squared A) [since 1 - cos squared A = sin squared A]
= (1 - cos A) / sin A
= 1/sin A - cos A/sin A
= cosec A - cot A
= RHS
Q6. Prove that (tan theta + sin theta) / (tan theta minus sin theta) = (sec theta + 1) / (sec theta minus 1).
LHS = (tan theta + sin theta) / (tan theta - sin theta)
Factor sin theta from numerator and denominator:
= sin theta (1/cos theta + 1) / [sin theta (1/cos theta - 1)]
Cancel sin theta:
= (sec theta + 1) / (sec theta - 1)
= RHS
Q7. Prove that sin theta / (1 + cos theta) + sin theta / (1 minus cos theta) = 2 cosec theta.
LHS = sin theta / (1 + cos theta) + sin theta / (1 - cos theta)
Take common denominator:
= [sin theta(1 - cos theta) + sin theta(1 + cos theta)] / [(1 + cos theta)(1 - cos theta)]
= [sin theta - sin theta cos theta + sin theta + sin theta cos theta] / (1 - cos squared theta)
= 2 sin theta / sin squared theta [since 1 - cos squared theta = sin squared theta]
= 2 / sin theta
= 2 cosec theta
= RHS
Q8. If sin theta = 4/5, find the value of (sec squared theta minus 1) / (1 minus cosec squared theta).
Given: sin theta = 4/5
So Opposite = 4, Hypotenuse = 5
Adjacent = sqrt(25 - 16) = sqrt(9) = 3
cos theta = 3/5, tan theta = 4/3, cosec theta = 5/4, sec theta = 5/3
Now:
(sec squared theta - 1) / (1 - cosec squared theta)
Numerator: sec squared theta - 1 = tan squared theta = 16/9
Denominator: 1 - cosec squared theta = -(cosec squared theta - 1) = -cot squared theta = -9/16
= (16/9) / (-9/16)
= (16/9) x (-16/9)
= -256/81
Exercise 6.3 — Heights and Distances
Q1. A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60 degrees. Find the height of the tower.
Let the height of the tower = h metres.
Distance from the foot = 20 m
Angle of elevation = 60 degrees
In the right triangle:
tan 60 degrees = h / 20
sqrt(3) = h / 20
h = 20 sqrt(3)
h = 20 x 1.732
h = 34.64 m
Q2. A ladder leaning against a wall makes an angle of 60 degrees with the ground. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.
Let length of the ladder = l metres.
Distance of foot from wall = 2.5 m
Angle with ground = 60 degrees
The ladder is the hypotenuse, distance from wall is the adjacent side.
cos 60 degrees = Adjacent / Hypotenuse
1/2 = 2.5 / l
l = 2.5 x 2
l = 5 m
Q3. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60 degrees. Find the length of the string, assuming there is no slack in the string.
Height of kite above ground = 60 m
Angle of inclination = 60 degrees
The height is the opposite side, the string is the hypotenuse.
sin 60 degrees = Opposite / Hypotenuse
sqrt(3)/2 = 60 / l
l = 60 x 2 / sqrt(3)
l = 120 / sqrt(3)
Rationalising:
l = (120 x sqrt(3)) / 3
l = 40 sqrt(3) m
l = 40 x 1.732
l = 69.28 m
Q4. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Determine the height of the tower.
Let the height of the tower = H metres.
Height of building = 7 m
Distance between building and tower = d metres.
From the angle of depression of 45 degrees (foot of tower):
tan 45 degrees = 7 / d
1 = 7 / d
d = 7 m
From the angle of elevation of 60 degrees (top of tower):
The height above the building top = H - 7
tan 60 degrees = (H - 7) / d
sqrt(3) = (H - 7) / 7
H - 7 = 7 sqrt(3)
H = 7 + 7 sqrt(3)
H = 7(1 + sqrt(3))
H = 7 x (1 + 1.732)
H = 7 x 2.732
H = 19.124 m
Q5. As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships approaching it are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Height of lighthouse = 75 m
For the nearer ship (angle of depression = 45 degrees):
tan 45 degrees = 75 / d1
1 = 75 / d1
d1 = 75 m
For the farther ship (angle of depression = 30 degrees):
tan 30 degrees = 75 / d2
1/sqrt(3) = 75 / d2
d2 = 75 sqrt(3) m
Distance between the two ships:
= d2 - d1
= 75 sqrt(3) - 75
= 75(sqrt(3) - 1)
= 75(1.732 - 1)
= 75 x 0.732
= 54.9 m
Q6. The angle of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Find the height of the tower.
Let height of tower = h metres.
From point at 4 m (angle of elevation = alpha):
tan alpha = h / 4 ... (1)
From point at 9 m (angle of elevation = 90 degrees - alpha):
tan(90 degrees - alpha) = h / 9
cot alpha = h / 9 ... (2)
Since tan alpha x cot alpha = 1:
(h/4) x (h/9) = 1
h squared / 36 = 1
h squared = 36
h = 6 m
Q7. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30 degrees with the ground. The distance between the foot of the tree and the point where the top touches the ground is 10 m. Find the height of the tree.
Let the broken part of tree = l metres.
Angle with ground = 30 degrees
Distance from foot to top on ground = 10 m
The broken part is the hypotenuse, the distance on ground is the adjacent side.
cos 30 degrees = 10 / l
sqrt(3)/2 = 10 / l
l = 20 / sqrt(3)
l = 20 sqrt(3) / 3 m (rationalised)
The standing part of tree (adjacent side in the triangle with angle 30 degrees):
tan 30 degrees = standing part / 10
1/sqrt(3) = standing part / 10
standing part = 10/sqrt(3) = 10 sqrt(3)/3 m
Total height of tree = standing part + broken part
= 10 sqrt(3)/3 + 20 sqrt(3)/3
= 30 sqrt(3)/3
= 10 sqrt(3) m
= 10 x 1.732
= 17.32 m
Exercise 6.4 — Multiple Choice Questions
Q1. The value of sin squared 30 degrees + cos squared 30 degrees is:
(a) 0 (b) 1/2 (c) 1 (d) 2
sin 30 degrees = 1/2, cos 30 degrees = sqrt(3)/2
sin squared 30 degrees + cos squared 30 degrees
= (1/2) squared + (sqrt(3)/2) squared
= 1/4 + 3/4
= 4/4
= 1
Also by the identity sin squared theta + cos squared theta = 1 for any theta.
Q2. If tan A = sqrt(3), then the value of A is:
(a) 30 degrees (b) 45 degrees (c) 60 degrees (d) 90 degrees
tan A = sqrt(3)
From the standard angle values table:
tan 60 degrees = sqrt(3)
Therefore A = 60 degrees.
Q3. The value of 2 tan squared 45 degrees + cos squared 30 degrees minus sin squared 60 degrees is:
(a) 0 (b) 1 (c) 2 (d) 1/2
tan 45 degrees = 1
cos 30 degrees = sqrt(3)/2
sin 60 degrees = sqrt(3)/2
= 2(1) squared + (sqrt(3)/2) squared - (sqrt(3)/2) squared
= 2 + 3/4 - 3/4
= 2
Q4. If sin theta = 1/3, then cos theta equals:
(a) 2/3 (b) 2 sqrt(2)/3 (c) sqrt(2)/3 (d) 1/3
Given: sin theta = 1/3
Using identity: cos squared theta = 1 - sin squared theta
= 1 - 1/9
= 8/9
cos theta = sqrt(8/9) = 2 sqrt(2) / 3
Q5. The angle of elevation of the top of a tower from a point 100 m away from its base is 45 degrees. The height of the tower is:
(a) 50 m (b) 100 m (c) 50 sqrt(2) m (d) 100 sqrt(2) m
Let height = h.
Distance = 100 m
Angle of elevation = 45 degrees
tan 45 degrees = h / 100
1 = h / 100
h = 100 m
Q6. If sin(A + B) = 1 and sin(A minus B) = 1/2, then A and B are:
(a) A = 45 degrees, B = 15 degrees (b) A = 60 degrees, B = 30 degrees (c) A = 75 degrees, B = 15 degrees (d) A = 30 degrees, B = 60 degrees
sin(A + B) = 1 => A + B = 90 degrees
sin(A - B) = 1/2 => A - B = 30 degrees
Adding: 2A = 120 degrees => A = 60 degrees
Subtracting: 2B = 60 degrees => B = 30 degrees
Q7. The value of (sin 30 degrees + cos 30 degrees) squared is:
(a) 1 (b) 2 (c) 1 + sqrt(3)/2 (d) 1/2
sin 30 degrees = 1/2, cos 30 degrees = sqrt(3)/2
= (1/2 + sqrt(3)/2) squared
= ((1 + sqrt(3))/2) squared
= (1 + 2 sqrt(3) + 3) / 4
= (4 + 2 sqrt(3)) / 4
= 1 + sqrt(3)/2
Q8. If the shadow of a tower is sqrt(3) times its height, then the angle of elevation of the sun is:
(a) 30 degrees (b) 45 degrees (c) 60 degrees (d) 90 degrees
Let height of tower = h
Shadow = sqrt(3) x h
Angle of elevation = theta
tan theta = height / shadow = h / (sqrt(3) x h) = 1/sqrt(3)
tan 30 degrees = 1/sqrt(3)
Therefore theta = 30 degrees.
Common Mistakes to Avoid
1. Confusing opposite and adjacent sides. Always label the triangle first. Opposite is across from the angle, adjacent is next to the angle (not the hypotenuse). 2. Starting identity proofs from both sides simultaneously. Always pick one side (preferably the more complex one) and simplify step by step to reach the other side. 3. Forgetting to rationalise denominators. When your answer has a square root in the denominator (like 1/sqrt(3)), multiply top and bottom by the root to rationalise. 4. Using the wrong trigonometric ratio in word problems. Before reaching for a formula, draw the triangle, mark what you know and what you need, then choose the ratio that connects them directly. 5. Mixing up angle of elevation and angle of depression. Elevation goes up from horizontal; depression goes down. But the angle of depression from the top equals the angle of elevation from the bottom (alternate interior angles). 6. Omitting units in the final answer. Heights and distances must include metres or the appropriate unit. A bare number loses marks. 7. Not using the Pythagorean theorem correctly. When one ratio is given, set up the right triangle and use hypotenuse squared = opposite squared + adjacent squared, not addition of sides. 8. Applying identities with wrong signs. sin squared plus cos squared equals 1, not sin squared minus cos squared. Write the identity fully before substituting. 9. Drawing the diagram after solving. Always sketch the diagram first. It prevents wrong ratio selection and catches misread problems early. 10. Ignoring the complementary angle shortcut. When angles of elevation from two points are complementary, use the tan alpha x tan(90 degrees minus alpha) = 1 trick for faster solutions.
Exam Checklist
Before your exam, confirm each item:
- [ ] All six trigonometric ratios memorised with their side definitions
- [ ] Standard angle values table for 0, 30, 45, 60, 90 degrees memorised
- [ ] Three fundamental identities can be written from memory without error
- [ ] Complementary angle relationships memorised (sin(90 minus theta) = cos theta, etc.)
- [ ] Given one ratio, I can find all other five ratios by drawing a right triangle
- [ ] Identity proofs: I always start from one side and simplify step by step
- [ ] Identity proofs: I factorise and take common factors before expanding
- [ ] Heights and distances: I draw a labelled diagram before writing any equation
- [ ] Heights and distances: I mark angle of elevation or depression correctly
- [ ] Heights and distances: I choose the ratio that directly links known and unknown
- [ ] Heights and distances: I include correct units in the final answer
- [ ] Heights and distances: I handle combined elevation and depression problems by finding the horizontal distance first
- [ ] Rationalisation: I rationalise denominators in the final answer
- [ ] MCQs: I verify my answer by substituting back when time permits
Frequently Asked Questions
Draw a right triangle, mark angle, opposite side, adjacent side, and hypotenuse before choosing a ratio.