Introduction to Trigonometry and Circles. For each chapter: the theory you need, a step-by-step method to turn the words of a question into a labelled diagram, fully worked examples, and practice questions with answers.
Every Chapter 8 question is about a right-angled triangle. The six trigonometric ratios compare its sides, but the names of two sides depend on which angle you are looking from.
Right angle at B. Sides are named with respect to ∠A.
Side
How to find it (for ∠A)
In ΔABC
Hypotenuse
Faces the right angle. Always the longest side. Never changes.
AC
Opposite (perpendicular)
Faces the angle you are using — does not touch it.
BC
Adjacent (base)
Touches the angle, but is not the hypotenuse.
AB
sin A = Opposite / Hypotenuse = BC/AC
cos A = Adjacent / Hypotenuse = AB/AC
tan A = Opposite / Adjacent = BC/AB
cosec A = 1/sin A · sec A = 1/cos A · cot A = 1/tan A
Memory line: Some People Have · Curly Brown Hair · Through Proper Brushing → sin = P/H, cos = B/H, tan = P/B
2Method: from question to figure in 6 steps
Find the right angle. Read which triangle and where the 90° is (“ΔPQR right-angled at Q”). If the question only says “sin A = …”, use the NCERT standard: ΔABC right-angled at B.
Draw the right angle first at the bottom-right corner and mark it with a small square. This fixes the hypotenuse.
Mark the given angle with a small arc at the bottom-left corner.
Label the sides in this order: hypotenuse (opposite the square) → opposite (faces the arc) → adjacent (the one left over).
Turn the ratio into sides using k. sin A = 3/5 means BC : AC = 3 : 5, not that BC is exactly 3. So write BC = 3k, AC = 5k. The k cancels in every ratio.
Find the missing side with Pythagoras (H² = P² + B²), write it on the figure, then read any ratio straight off the diagram.
Key idea: if the question switches to the other acute angle (∠C instead of ∠A), the hypotenuse stays the same but opposite and adjacent swap. Re-label before you write any ratio.
3Worked examples
Example 1
Given sin A = 3/4, calculate cos A and tan A.
sin A = 3/4 → BC = 3k, AC = 4k, AB found by Pythagoras
1
No triangle named, so draw ΔABC right-angled at B. Mark ∠A at the bottom-left.
2
sin A = Opposite/Hypotenuse = BC/AC = 3/4. So label BC = 3k and AC = 4k.
3
Find AB by Pythagoras: AB² = AC² − BC² = 16k² − 9k² = 7k², so AB = √7 k. Write it on the base.
4
Read from the figure: cos A = AB/AC = √7k / 4k = √7/4
5
tan A = BC/AB = 3k / √7k = 3/√7
Answer: cos A = √7/4, tan A = 3/√7
Teaching tip: Ask the student to say which side is the hypotenuse before writing anything. Then check: is the hypotenuse (4k) the longest side? 4 > 3 > 2.65 ✓.
Example 2
If 15 cot A = 8, find sin A and sec A.
cot A = 8/15 → adjacent AB = 8k, opposite BC = 15k
1
First rewrite the condition as a ratio: cot A = 8/15.
2
cot A = Adjacent/Opposite = AB/BC. So AB = 8k (base) and BC = 15k (perpendicular).
3
Hypotenuse: AC² = (8k)² + (15k)² = 64k² + 225k² = 289k², so AC = 17k.
4
sin A = BC/AC = 15k/17k = 15/17
5
sec A = AC/AB = 17k/8k = 17/8
Answer: sin A = 15/17, sec A = 17/8
Teaching tip: cot is the trap here — students put 8 on the opposite side because 8 is on top. Remind them: cot = 1/tan = Base/Perpendicular.
Example 3
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.
Unknown side = x, so the other = 25 − x
1
The right angle is at Q, so Q goes at the bottom-right corner. ∠P is at the bottom-left, R at the top.
2
PR is opposite the right angle → hypotenuse. QR faces ∠P → opposite. PQ = 5 cm → adjacent.
3
Two sides are unknown but linked. Let QR = x; then PR = 25 − x. Write both on the figure.
So QR = 12 cm, PR = 13 cm. sin P = 12/13, cos P = 5/13, tan P = 12/5.
Answer: sin P = 12/13, cos P = 5/13, tan P = 12/5
Teaching tip: This is the clearest case where the figure solves the problem: without the sketch, students cannot tell which side is the hypotenuse. Let the student draw it before you say anything.
Example 4
In ΔABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find (i) sin A, cos A (ii) sin C, cos C.
Same triangle, two different angles — opposite and adjacent swap
1
Hypotenuse: AC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm.
2
Looking from ∠A: opposite = BC = 7, adjacent = AB = 24.
3
sin A = 7/25, cos A = 24/25.
4
Now look from ∠C (the top vertex): opposite = AB = 24, adjacent = BC = 7. The hypotenuse is still AC.
5
sin C = 24/25, cos C = 7/25.
Answer: sin A = 7/25, cos A = 24/25; sin C = 24/25, cos C = 7/25
Teaching tip: Point out that sin A = cos C. This is because A + C = 90° — a preview of complementary angles: sin(90° − θ) = cos θ.
Common mistakes in Chapter 8 figures
Putting the right-angle square at the angle that is given (∠A). The given angle is always one of the two acute angles.
Writing the ratio numbers as exact lengths. Use 3k, 5k — it shows the examiner you know a ratio gives proportion only.
Calling the wrong side the hypotenuse. Check it is the longest side, and that sin and cos come out ≤ 1.
Using the same “opposite” for a different angle. Re-label every time the angle changes.
Chapter 10Circles
4Theory: tangents
A line and a circle can meet in three ways. The tangent is the line that touches the circle at exactly one point, called the point of contact.
A tangent touches the circle at exactly one point — the point of contact P
Position of the point
Number of tangents
Inside the circle
0
On the circle
1
Outside the circle
2
Theorem 10.1 — Tangent ⟂ radius
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
OP ⟂ tangent. Any other point Q on the tangent is farther: OQ > OP
Why: every point Q on the tangent except P lies outside the circle, so OQ > OP. OP is therefore the shortest distance from O to the line, and the shortest distance from a point to a line is the perpendicular.
Theorem 10.2 — Tangents from an external point are equal
The standard figure: two tangents, two radii, one line OP. PA = PB (tick marks)
Proof (draw this figure first):
1
Join OA, OB and OP.
2
In ΔOAP and ΔOBP: OA = OB (radii of the same circle).
So ΔOAP ≅ ΔOBP by RHS congruence, and therefore PA = PB (CPCT).
Results you get from this figure
PA = PB — equal tangents
∠APO = ∠BPO — OP bisects the angle between the tangents
∠AOP = ∠BOP
∠APB + ∠AOB = 180° — angles of quadrilateral OAPB add to 360°, and two of them are 90°
5Method: drawing any circle–tangent figure
Draw the circle and mark the centre O with a dot. Even if the question doesn’t mention the centre, you will need it.
Place the external point P to the right, about 2–3 times the radius away, so the tangents are not cramped.
Draw the tangents with a ruler, just grazing the circle — they must not cut through it. Name the points of contact (A, B).
Join the centre to each point of contact and mark the right angles straight away. This is the hidden step: almost every question is solved through these right triangles.
Join OP if the question involves the distance from the centre or angles at O or P.
Write every given value on the figure and mark equal tangents with matching tick marks.
Polygon around a circle (triangle or quadrilateral whose sides all touch the circle): name every point of contact, then from each corner pair up the two tangent pieces — they are equal. Colour each pair the same, as in Example 9.
6Worked examples
Example 5
The length of a tangent from a point Q to a circle is 24 cm, and the distance of Q from the centre is 25 cm. Find the radius of the circle.
Join OP (radius) — the right angle at P gives a right triangle OPQ
1
Draw the circle with centre O, the external point Q, and one tangent QP touching at P.
2
Join OP (radius) and mark ∠OPQ = 90° (tangent ⟂ radius). Join OQ.
3
Write the values: PQ = 24 cm, OQ = 25 cm. OQ is opposite the right angle, so it is the hypotenuse.
Teaching tip: The whole question is solved by one line the question never mentions: the radius OP. Train the student to draw it every time without being told.
Example 6
Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Chord AB of the big circle touches the small circle at P
1
Concentric = same centre O. Draw the small circle inside the big one.
2
Draw chord AB of the big circle so that it just touches the small circle at P. AB is a tangent to the small circle.
3
Join OP (radius of small circle, 3 cm) → OP ⟂ AB. Join OA (radius of big circle, 5 cm).
4
In right ΔOPA: AP² = OA² − OP² = 25 − 9 = 16, so AP = 4 cm.
5
The perpendicular from the centre bisects the chord, so AB = 2 × AP = 8 cm.
Answer: AB = 8 cm
Teaching tip: Students often join OA to the wrong circle. Colour the two circles differently on the board so it is obvious OP = 3 and OA = 5.
Example 7
PA and PB are tangents from an external point P to a circle with centre O. If ∠APB = 80°, find ∠POA.
Join OA, OB (right angles) and OP — find ∠POA
1
Draw the standard figure: circle, centre O, external point P, tangents PA and PB.
ΔOAP ≅ ΔOBP (Theorem 10.2 proof), so OP bisects ∠AOB: ∠POA = 100° ÷ 2 = 50°.
5
Check using ΔOAP: ∠OPA = 40°, ∠OAP = 90°, so ∠POA = 180° − 130° = 50° ✓
Answer: ∠POA = 50°
Teaching tip: Show both routes (quadrilateral and triangle). Getting the same answer twice is the best confidence-builder before an exam.
Example 8
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
AB is a diameter; PQ and RS are the tangents at its ends
1
Draw the circle with centre O and a diameter AB (vertical is easiest).
2
Draw tangent PQ at A and tangent RS at B. These are the two lines we must prove parallel.
3
OA is a radius and PQ is the tangent at A, so ∠OAQ = 90° — that is, ∠BAQ = 90°.
4
Similarly at B: ∠ABR = 90°.
5
∠BAQ and ∠ABR are alternate interior angles made by transversal AB, and they are equal (both 90°). So PQ ∥ RS.
Answer: Proved: PQ ∥ RS
Teaching tip: In proofs, the figure plus the line “OA ⟂ PQ (tangent ⟂ radius)” usually carries most of the marks. Make the student write the reason in brackets each time.
Example 9
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Same colour = equal tangents from the same corner
1
Draw the circle first, then draw four tangent lines around it so they form ABCD. Every side must touch the circle.
2
Name the points of contact: P on AB, Q on BC, R on CD, S on DA.
3
From each corner the two tangents are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
4
Add all four: AP + BP + CR + DR = AS + BQ + CQ + DS.
5
Group the left side as (AP + PB) + (CR + RD) = AB + CD and the right side as (AS + SD) + (BQ + QC) = AD + BC.
Answer: AB + CD = AD + BC (Proved)
Teaching tip: The colouring is the whole trick. Once each corner’s pair is the same colour, students can see that each side is made of two different colours, and the sum works out.
Common mistakes in Chapter 10 figures
Not joining the radius to the point of contact — the right angle is where the solution comes from.
Drawing a “tangent” that cuts the circle. Use a ruler and let it just touch.
Forgetting to mark the centre, or labelling it in the wrong place on concentric circles.
Assuming things that only look true. Only use proven results: tangent ⟂ radius, equal tangents, OP bisects ∠APB.
In polygon questions, pairing tangents from different corners. Each pair must start from the same corner.
PracticeDraw the figure first, then solve
7Chapter 8 — Introduction to Trigonometry
Practice 1
If tan A = 4/3, find sin A and cos A.
Show answer
Opposite BC = 4k, adjacent AB = 3k, hypotenuse AC = 5k. sin A = 4/5, cos A = 3/5.
Practice 2
If sec θ = 13/12, find all the other trigonometric ratios.
Show answer
sec = H/B → hypotenuse 13k, base 12k, perpendicular = √(169 − 144) k = 5k. sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5.
Practice 3
In ΔABC, right-angled at B, if tan A = 1/√3, find the value of sin A cos C + cos A sin C.
Show answer
BC = k, AB = √3 k, AC = 2k. sin A = 1/2, cos A = √3/2. From ∠C: sin C = AB/AC = √3/2, cos C = BC/AC = 1/2. Value = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
Practice 4
If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A.
Show answer
Base 4k, perpendicular 3k, hypotenuse 5k. tan A = 3/4. LHS = (1 − 9/16)/(1 + 9/16) = 7/25. RHS = 16/25 − 9/25 = 7/25. Yes, they are equal.
Practice 5
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Find ∠QPR and ∠PRQ.
Show answer
From ∠P: adjacent PQ = 3, hypotenuse PR = 6, so cos P = 1/2 → ∠QPR = 60°, and ∠PRQ = 30°.
Practice 6
In ΔABC, right-angled at C, AC = 5 cm and BC = 12 cm. Find sin A and cos B.
Show answer
Careful — the right angle is at C, so AB is the hypotenuse: AB = 13 cm. sin A = BC/AB = 12/13; cos B = BC/AB = 12/13.
8Chapter 10 — Circles
Practice 7
From a point P, 13 cm from the centre of a circle, the length of the tangent is 12 cm. Find the radius.
Show answer
Radius ⟂ tangent, so r² = 13² − 12² = 25 → r = 5 cm.
Practice 8
Two concentric circles have radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle that touches the smaller circle.
Show answer
Half chord = √(6.5² − 2.5²) = √(42.25 − 6.25) = √36 = 6 cm. Chord = 12 cm.
Practice 9
PA and PB are tangents from P to a circle with centre O. If ∠AOB = 110°, find ∠APB.
Show answer
∠APB + ∠AOB = 180° → ∠APB = 70°.
Practice 10
A circle is inscribed in ΔABC, touching AB, BC and CA at P, Q and R. If AB = 10 cm, BC = 8 cm and CA = 12 cm, find AP, BQ and CR.
Show answer
Let AP = AR = x, BP = BQ = y, CQ = CR = z. Then x + y = 10, y + z = 8, z + x = 12. Adding: x + y + z = 15. So AP = 7 cm, BQ = 3 cm, CR = 5 cm.
Practice 11
A quadrilateral ABCD circumscribes a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, find AD.
Show answer
AB + CD = AD + BC → 6 + 4 = AD + 7 → AD = 3 cm.
Practice 12
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at Q so that OQ = 12 cm. Find PQ.
Show answer
OP ⟂ PQ, so PQ² = 12² − 5² = 144 − 25 = 119 → PQ = √119 cm.
Practice 13
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended at the centre by the line segment joining the points of contact.
Show answer
Draw tangents PA, PB and join OA, OB. ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, the angles add to 360°, so ∠APB + ∠AOB = 360° − 180° = 180°. Hence they are supplementary.
