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Example 1 - Given tan A = 4/3, find other ratios - Examples

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise
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Example 1 Given tan A = 4/3 , find the other trigonometric ratios of the angle A Given, tan A = 4/3 (๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด)/(๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด) = 4/3 ๐ต๐ถ/๐ด๐ต = 4/3 Let BC = 4x AB = 3x We find AC using Pythagoras Theorem In right triangle ABC Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (AC)2 = (BC)2 + (AB)2 (AC)2 = (4x)2 + (3x)2 (AC)2 = 16x2 + 9x2 (AC)2 = 25x2 AC = โˆš25๐‘ฅ2 AC = 5x Now, 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 = 4๐‘ฅ/5๐‘ฅ Sin A = 4/5 Similarly, cos A = (๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐ด)/๐ป๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’ cos A = ๐ด๐ต/๐ด๐ถ cos A = 3๐‘ฅ/5๐‘ฅ cos A = 3/5 Given, tan A=4/3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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