 ## 9.4.5 Solution of Triangles, Long Questions (Question 9)

Question 9 Diagram 6 shows a quadrilateral KLMN.Calculate (a) ∠KML, [2 marks] (b) the length, in cm, of KM,   [3 marks] (c) area, in cm2, of triangle KMN,   [3 marks](d) a triangle K’L’M’ has the same measurements as those given for triangle KLM, that is K’L’= 12.4 cm, L’M’= 9.5 cm and ∠L’K’M’= 43.2o, which is … Read more

## 9.4.4 Solution of Triangles, Long Questions (Question 7 & 8)

Question 7:Diagram below shows a quadrilateral ABCD where the sides AB and CD are parallel. ∠BAC is an obtuse angle.Given that AB = 14 cm, BC = 27 cm, ∠ACB = 30o and AB : DC = 7 : 3.Calculate(a) ∠BAC.(b) the length, in cm, of diagonal BD.(c) the area, in cm2, of quadrilateral ABCD.Solution: … Read more

## 9.4.3 Solution of Triangles Long Questions (Question 5 & 6)

Question 5: The diagram below shows a triangle ABC. (a) Calculate the length, in cm, of AC. (b) A quadrilateral ABCD is now formed so that AC is a diagonal, ∠ACD = 45° and AD = 14 cm. Calculate the two possible values of ∠ADC. (c) By using the acute ∠ADC from (b), calculate  (i) the length, … Read more

## 9.4.2 Solution of Triangles Long Questions (Question 3 & 4)

Question 3: The diagram shows a trapezium PQRS. PS is parallel to QR and QRS is obtuse. Find (a) the length, in cm, of QS, (b) the length, in cm, of RS, (c) ∠QRS, (d) the area, in cm2, of triangle QRS. Solution: (a) Q S sin P = P S sin Q Q S sin … Read more

## 9.4.1 Solution of Triangles Long Questions (Question 1 & 2)

Question 1: The diagram shows a quadrilateral ABCD. The area of triangle BCD is 12 cm2 and ∠BCD is acute. Calculate (a) ∠BCD, (b) the length, in cm, of BD, (c) ∠ABD, (d) the area, in cm2, quadrilateral ABCD. Solution: (a) Given area of triangle BCD = 12 cm2 ½ (BC)(CD) sin C = 12 ½ (7) … Read more

## 9.3 Area of Triangle

9.3 Area of Triangle Example: Calculate the area of the triangle above. Solution: A r e a = 1 2 a b sin C A r e a = 1 2 ( 7 ) ( 4 ) sin 40 o A r e a = 9  cm 2 Example:Diagram above shows a triangle ABC, where AC … Read more

## 9.2 The Cosine Rule

9.2 The Cosine Rule The cosine rule can be used when (i) two sides and the included angle, or (ii) three sides of a triangle are given. (A) If you know 2 sides and 1 angle between them [included angle] ⇒ Cosine rule Example: Calculate the length of AC, x, in cm for the triangle above. … Read more

## 9.1 The Sine Rule

9.1 The Sine RuleIn a triangle ABC in which the sides BC, CA and AB are denoted by a, b, and c as shown, and A, B, C are used to denote the angles at the vertices A, B, C respectively, The sine rule can be used when (i) two sides and one non-included angle … Read more

## SPM Form 4 Additional Mathematics Chapter 10 – Solution of Triangles

10 Solution of Triangles Sine Rule Cosine Rule Areas of Triangles Further Practice (Long Questions) Example 1 Example 2 Example 3 Example 4 Mind Map

## Sine Rule – Ambiguous Case – Example

ExampleBy taking into consideration the ambiguous case. Solve ∆ABC such that c = 6cm, a = 5cm and ∠A = 40°Answer:Step 1: To prove whether this is an ambiguous case/non-ambiguous case[begin{gathered}sin {40^o} = frac{y}{6} hfill \y = 6sin {40^o} hfill \y = 3.857cm hfill \end{gathered} ]y < a < cTherefore, this is an ambiguous case. … Read more