Day 103 – Cosine rules

Introduction We have discussed how to solve a triangle using the Pythagorean theorem. However, we have been solving right triangle alone, and trigonometric ratios may not be applicable in solving a non-right triangle. In this lesson, we will use the idea of the Pythagorean theorem and trigonometric ratios to derive cosines rule and use this rule to solve non- right angle triangles.

Vocabulary Acute angle This is an angle which is less than 90 ° in size. Obtuse angle This is an angle which is greater than 90 ° and less than 180 ° in size.

Consider the triangles below Consider the triangles below. In ∆𝐵𝐷𝐶, cos 𝐶= 𝑥 𝑎 ⟹𝑥=𝑎𝑐𝑜𝑠𝐶…(𝑖) 𝑎 2 = ℎ 2 + 𝑥 2 In ∆𝐴𝐵𝐷, 𝑐 2 = ℎ 2 + 𝑏−𝑥 2 ( By Pythagorean theorem) 𝑐 2 = ℎ 2 + 𝑏 2 −2𝑏𝑥+ 𝑥 2 𝑐 2 = 𝒉 𝟐 + 𝒙 𝟐 + 𝑏 2 −2𝑏𝑥 But ℎ 2 + 𝑥 2 =𝑎 2 Therefore, 𝑐 2 = 𝑎 2 + 𝑏 2 −2𝑏𝑥 … (𝑖𝑖) 𝑏−𝑥 c 𝑥 b h a C D A B

But 𝑥=𝑎𝑐𝑜𝑠𝐶…(𝑖) Substituting (𝑖) in 𝑖𝑖 we have, 𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 −𝟐𝒂𝒃𝒄𝒐𝒔𝑪 … 𝑖𝑖𝑖 Equation 𝑖𝑖𝑖 is referred to as cosine rule. ∠𝐶 is acute. Cosine rule is also true when the reference angle, ∠𝐶 ,is obtuse. To prove cosine rule when the reference angle is obtuse, we consider the following triangle.

DB is perpendicular to AB DB is perpendicular to AB. 𝑐 2 = ℎ 2 + 𝑥+𝑏 2 (By Pythagorean theorem) 𝑐 2 = 𝒉 𝟐 + 𝒙 𝟐 +2𝑏𝑥+ 𝑏 2 𝑐 2 = 𝑎 2 +2𝑥𝑏+ 𝑏 2 …(𝑖𝑖𝑖) (since 𝑎 2 = ℎ 2 + 𝑥 2 ) In ∆𝐵𝐶𝐷 cos 𝐶= 𝑥 𝑎 ∠𝐵𝐶𝐷 is supplementary to ∠𝐵𝐶𝐴, thus in ∆𝐵𝐶𝐴, cos 𝐶=− 𝑥 𝑎 ⟹𝑥=− acos 𝐶 h C c B b a A D 𝑥

Substituting this value of 𝑥 in (𝑖𝑖𝑖) we get, 𝑐 2 = 𝑎 2 −2𝑎𝑏 cos 𝐶 + 𝑏 2 𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 −𝟐𝒂𝒃 𝒄𝒐𝒔 𝑪 If we make cos 𝐶 subject of the formula we get, cos 𝐶= 𝑐 2 −( 𝑎 2 + 𝑏 2 ) 2𝑎𝑏 Cosine rule applies in all sides in a triangle. That is, 𝒃 𝟐 = 𝒄 𝟐 + 𝒂 𝟐 −𝟐𝒂𝒄 𝒄𝒐𝒔 𝑩 𝒂 𝟐 = 𝒄 𝟐 + 𝒃 𝟐 −𝟐𝒃𝒄 𝒄𝒐𝒔 𝑨 Cosine rule can only be used when we know all the sides or when we know one side and at least two sides in a triangle.

Example 1 Find the length of KL in the triangle below. J L 7 𝑖𝑛 4 𝑖𝑛 37 °

Solution Using cosine rule, 𝑗 2 = 𝑙 2 + 𝑘 2 −2𝑙𝑘𝑐𝑜𝑠 𝐽 𝑗 2 = 4 2 + 7 2 −2 ×7×4 𝑐𝑜𝑠 37 ° =16+49−44.72 = 20.28 𝑗= 20.28 = 4.5 𝑖𝑛

Example 2 Determine the size of ∠𝑋 15 𝑖𝑛 9 𝑖𝑛 10 𝑖𝑛 X Z Y

Solution Let ∠𝑋 be 𝜃 15 2 = 10 2 + 9 2 −2×10×9× cos 𝜃 225=181−180 cos 𝜃 225−181=−180 cos 𝜃 −180 cos 𝜃 =44 cos 𝜃 = 44 −180 =−0.24444 𝜃= 𝑐𝑜𝑠 −1 0.2444 = 104.1 °

homework Find the length of side BC C A B 85 ° 22 𝑖𝑛 19 𝑖𝑛

Answers to homework 27.8 𝑖𝑛

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