Elementary Mathematics

Elementary MathematicsChapter 10: TrigonometryCosine Rule : 

Elementary MathematicsChapter 10: TrigonometryCosine Rule

Slide 2: 

Cosine Rule: a2 = b2 + c2 – 2bc(cos A) b2 = a2 + c2 – 2ac(cos B) c2 = a2 + b2 – 2ab(cos C) Note: Cosine rule is used when two sides and an included angle are given.

Slide 3: 

Proof: When angle A is acute x = c(cos A), h2 = c2 - x2 By Pythagoras Theorem: a2 = (b - x)2 + h2 a2 = (b - x)2 + (c2 - x2) a2 = b2 – 2bx + x2 + c2 - x2 a2 = b2 – 2bx + x2 + c2 - x2 a2 = b2 – 2bx + c2 a2 = b2 + c2 – 2bx a2 = b2 + c2 – 2bc(cos A) A C B c b a x (b – x) h

Slide 4: 

Proof: When angle A is obtuse (180-A) h x A b C a c B x = c cos(180° - A) = c(-cos A) = - c cos A h2 = c2 - x2 By Pythagoras Theorem: a2 = (b + x)2 + h2 a2 = (b + x)2 + (c2 - x2) a2 = b2 + 2bx + x2 + c2 - x2 a2 = b2 + 2bx + c2 a2 = b2 + 2b(- c cos A) + c2 a2 = b2 + c2 – 2bc(cos A) A

Slide 5: 

When angle A is right angle: A C a c B b a2 = b2 + c2 – 2bc(cos 90°) a2 = b2 + c2 (Pythagoras Theorem!)

Slide 6: 

By considering angle B and C: a2 = b2 + c2 – 2bc(cos A) b2 = a2 + c2 – 2ac(cos B) c2 = a2 + b2 – 2ab(cos C) Note: Cosine rule is used when two sides and an included angle are given. Note:

Slide 7: 

Given triangle PQR, p = 5cm, q = 7cm, and R = 60°, find r. Solution: r2 = p2 + q2 – 2pq(cos R) = 52 + 72 – 2(5)(7)(cos 60°) = 39 r = 6.24 cm (to 3 sig figs) Exercise 10D, Question1

Slide 8: 

Ex 10D, Question Q6: Given triangle ABC, a = 3.8 cm, b = 5.3 cm,c =6.7 cm find the largest angle. Thinking Process: Largest angle is opposite longest length. Thus, largest angle is angle C. Solution: c2 = a2 + b2 – 2ab(cos C) (6.7)2 = (3.8)2 + (5.3)2 – 2(3.8)(5.3)(cos C) cos C = (3.8)2 + (5.3)2 – (6.7)2 2(3.8)(5.3) C = cos-1 (-0.05859) = (180° - 86.64°) = 93.4° (to 1 dec plc) Cosine negative, angle C is in the 2nd quadrant!

Slide 9: 

Homework: Page 222, Ex 10D, Q13, 14, 15, 16 Do in CCHMS foolscap.