Application of Trigonometric Ratios to Problem Solving – SS1 Mathematics Lesson Note (Second Term)

Trigonometric ratios are fixed comparisons between the sides of a right-angled triangle, used to find unknown lengths and angles.

Behavioural objectives

By the end of the lesson, students should be able to:

1. Write the sine, cosine and tangent of the acute angles of a right-angled triangle.
2. Use the relationship between sin θ and cos (90° – θ).
3. Use trigonometric ratios to solve problems on distances and angles.
4. State Pythagoras’ theorem.
5. Find the length of a chord of a circle using trigonometric ratios or Pythagoras’ theorem.

What are the sine, cosine and tangent of an acute angle?

The sine, cosine and tangent of an acute angle are ratios of two sides of a right-angled triangle.

• Hypotenuse: the longest side, opposite the right angle.
• Opposite: the side facing the angle θ.
• Adjacent: the side next to θ that is not the hypotenuse.

Soh cah toa

SOH CAH TOA is a memory aid for the three ratios.

• SOH: sin θ = Opp/Hyp.
• CAH: cos θ = Adj/Hyp.
• TOA: tan θ = Opp/Adj.

What are the ratios of the angles in triangle ABC?

Triangle ABC is right-angled at A, so side a is the hypotenuse.

• Side a: opposite angle A, the hypotenuse.
• Side b: opposite angle B.
• Side c: opposite angle C.

What is the relationship between sin θ and cos (90° – θ)?

The sine of an angle equals the cosine of its complement.

• Complementary angles: B and C add up to 90°, so if B = θ then C = 90° – θ.
• Sine rule of complements: sin θ = cos (90° – θ) = b/a.
• Cosine rule of complements: cos θ = sin (90° – θ) = c/a.

How do we solve problems using trigonometric ratios?

Choose the ratio that links the side or angle you want with the one you already know.

1. Draw a clear sketch and mark the right angle.
2. Label the sides Opp, Adj and Hyp from the angle you are using.
3. Pick sine, cosine or tangent using SOH CAH TOA.
4. Write the equation with the numbers put in.
5. Solve for the unknown, using tables or a calculator.
6. Give the answer to the required accuracy, with units.

Example 1: finding an angle

A triangle has sides 8 cm and 5 cm and an angle of 90° between them. Calculate the smallest angle of the triangle.

1. The smallest angle is opposite the shortest side, 5 cm. Call it θ.
2. The side 8 cm is adjacent to θ.
3. tan θ = Opp/Adj = 5/8 = 0.625.
4. θ = 32° to the nearest degree.

Example 2: finding a distance

Town Y is 200 km from town X in a direction of 40°. How far is Y east of X?

1. From the diagram, the distance of Y east of X is ZY.
2. Use the right-angled triangle XZY, with XY = 200 km as the hypotenuse.
3. ZY is opposite the 40° angle, so sin 40° = ZY/200 km.
4. ZY = 200 km × sin 40°.
5. ZY = 200 km × 0.6428 = 128.56 km.
6. ZY = 128.6 km to 1 d.p.

What is Pythagoras’ theorem?

Pythagoras’ theorem states that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

• Formula: c² = a² + b², where c is the hypotenuse.
• Use: it finds a missing side when two sides are known and no angle is given.
• Example: sides 3 cm and 4 cm give a hypotenuse of 5 cm, since 9 + 16 = 25.

How do we find the length of a chord of a circle?

A chord is a straight line joining two points on a circle, and its length is found from a right-angled triangle drawn from the centre.

• Key fact: the perpendicular from the centre to a chord bisects the chord.
• Radius: the line from the centre to either end of the chord is a radius, and it is the hypotenuse of the triangle.
• Angles given: use sine, cosine or tangent.
• No angle given: use Pythagoras’ theorem.

Steps for finding a chord length

1. Draw the circle, the chord and the perpendicular from the centre.
2. Mark the radius as the hypotenuse of the right-angled triangle.
3. Use Pythagoras’ theorem to find half the chord.
4. Double it to get the whole chord.

Example 3

A chord is drawn 3 cm away from the centre of a circle of radius 5 cm. Calculate the length of the chord.

1. Sketch a right-angled triangle and label it. Let the half-chord be AB.
2. |AB|² + 3² = 5² (Pythagoras’ theorem).
3. |AB|² = 5² – 3² = 25 – 9 = 16.
4. AB = √16 = 4 cm.
5. B is the midpoint of chord AC, so AC = 2 × AB = 2 × 4 cm = 8 cm.

Example 4

O is the centre of circle HKL. HK = 16 cm and HL = 10 cm. The perpendicular from O to HK is 4 cm. Find the length of the perpendicular from O to HL.

1. Let the distance from O to HL be x cm.
2. The perpendicular from O to HK meets HK at its midpoint M, so HM = 8 cm.
3. In right-angled triangle OMH: |OH|² = |HM|² + |MO|² = 8² + 4².
4. |OH|² = 64 + 16 = 80, so |OH| = √80 cm.
5. OH is the radius, so r = OH = OL = √80 cm.
6. The perpendicular from O to HL meets HL at its midpoint N, so NL = 5 cm.
7. In right-angled triangle ONL: OL² = ON² + NL².
8. (√80)² = x² + 5², so 80 – 25 = x².
9. x² = 55, so x = √55 = 7.416 cm.
10. The perpendicular from O to HL is 7.416 cm.

What mistakes should students avoid?

Most errors come from labelling sides wrongly or rounding too early.

• Wrong labels: Opp and Adj change when you change the angle.
• Wrong hypotenuse: it is always opposite the right angle and is the longest side.
• Half chord: forgetting to double the half-chord at the end.
• Early rounding: keep full values until the last line.

Evaluation

1. In triangle ABC, right-angled at A, write down sin B, cos B and tan B in terms of the sides a, b and c.
2. If B = θ, write down C in terms of θ and show that sin θ = cos (90° – θ).
3. State Pythagoras’ theorem.
4. A ladder 20 m long rests against a vertical wall so that the foot of the ladder is 9 m from the wall.
   1. Find, correct to the nearest degree, the angle that the ladder makes with the wall.
   2. Find, correct to 1 d.p., the height above the ground at which the upper end of the ladder touches the wall. Use tables of trigonometric ratios.
5. A chord is 6 cm from the centre of a circle of radius 10 cm. Find the length of the chord.

Assignment

Town Y is 200 km from town X in a direction of 40°. Using the method in Example 2, find how far Y is north of X. Give your answer correct to 1 d.p.