Example 3 - Consider ACB, AB = 29, BC = 21 and angle ABC - Concept wise

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise
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Example 3 Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see fig.). Determine the values of (i) cos2 θ + sin2 θ, Step1 : Finding sides of triangle In right triangle ABC, using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AB2 = AC2 + BC2 AC2 = AB2 – BC2 AC2 = (29)2 – (21)2 Using a2 – b2 = (a + b) (a – b) AC2 = (29 – 21) (29 + 21) AC2 = (8) (50) AC = √(8×50) AC = √400 AC = 20 Step 2 : Finding sin θ , cos θ We have to find out , cos2 𝜃 + sin2 𝜃 Putting values = (21/29)^2+(20/29)^2 = ((21)2 + (20)2)/292 = (441 + 400)/841 = 841/841 = 1 So, cos2𝜃 + sin2 𝜃 = 1 Example 3 Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ(see fig.). Determine the values of (ii) cos2 θ – sin2 θ. cos2 𝜃 – sin2 𝜃 Putting values = (21/29)^2−(20/29)^2 = ((21)2 − (20)2)/292 Using a2 – b2 = (a + b) (a – b) = ((21 + 20)(21 − 20))/292 = ((41)(1))/841 = 41/841

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