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Solution of Triangles

9.4.5 Solution of Triangles, Long Questions (Question 9)

February 15, 2022November 9, 2020 by

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

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

January 21, 2022November 8, 2020 by

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

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

January 21, 2022November 7, 2020 by

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

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

January 21, 2022November 6, 2020 by

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

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

January 21, 2022November 5, 2020 by

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

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9.3 Area of Triangle

January 21, 2022October 31, 2020 by

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

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9.2 The Cosine Rule

January 21, 2022October 28, 2020 by

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

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9.1 The Sine Rule

January 21, 2022October 25, 2020 by

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

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SPM Form 4 Additional Mathematics Chapter 10 – Solution of Triangles

September 19, 2019November 6, 2010 by

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

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Sine Rule – Ambiguous Case – Example

September 20, 2019June 25, 2010 by

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

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