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Ex 8.1, 9 - In ABC, if tan A = 1/ root 3, find sin A cos C - Finding values of expressions

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise
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Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/โˆš3, find the value of sin A cos C + cos A sin C tan A = 1/โˆš3 (๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘ก๐‘œ ๐ด)/(๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐ด) = 1/โˆš3 ๐ต๐ถ/๐ด๐ต = 1/โˆš3 Let BC = x & AB = โˆš3 x We find AC using Pythagoras theorem Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AC2 = AB2 + BC2 AC2 = (โˆš3 ๐‘ฅ)^2+(๐‘ฅ)2 AC2 = 3x2 + x2 AC2 = 4x2 AC = โˆš4๐‘ฅ2 AC = 2x We need to find sin A , cos A , sin C & cos C sin A = (๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด)/๐ป๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’ sin A = ๐ต๐ถ/๐ด๐ถ sin A = ๐‘ฅ/2๐‘ฅ sin A = 1/2 cos A = (๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด )/๐ป๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’ cos A = ๐ด๐ต/๐ด๐ถ cos A = (โˆš3 ๐‘ฅ)/2๐‘ฅ cos A = โˆš3/2 We have to find out. sin A cos C + cos A sin C Putting sin A = 1/2 , cos A = โˆš3/2 , sin C = โˆš3/2 & cos C = 1/2 = (1/2)ร—(1/2)+(โˆš3/2)ร—(โˆš3/2) = 1/4 + (โˆš3 ร— โˆš3)/4 = 1/4 + 3/4 = (1 + 3)/4 = 4/4 = 1 So, sin A cos C + cos A sin C = 1 Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/โˆš3, find the value of (ii) cos A cos C โ€“ sin A sin C cos A cos C โ€“ sin A sin C Putting sin A = 1/2 , cos A = โˆš3/2 , sin C = โˆš3/2 & cos C = 1/2 = (โˆš3/2)ร—1/2โˆ’(1/2)ร—(โˆš3/2) = (โˆš3/4)โˆ’(โˆš3/4) = 0 Hence , cos A cos C โ€“ sin A sin C = 0

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