Example 1 - Given tan A = 4/3, find other ratios - Chapter 8 Class 10 - Examples

part 2 - Example 1 - Examples - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry
part 3 - Example 1 - Examples - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry

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Transcript

Example 1 Given tan A = 4/3 , find the other trigonometric ratios of the angle A Given, tan A = 4/3 (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’ 𝐴)/(𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’ 𝐴) = 4/3 𝑩π‘ͺ/𝑨𝑩 = πŸ’/πŸ‘ Let BC = 4x AB = 3x We find AC using Pythagoras Theorem In right triangle ABC Using Pythagoras theorem Hypotenuse2 = Height2 + Base2 (AC)2 = (BC)2 + (AB)2 (AC)2 = (4x)2 + (3x)2 (AC)2 = 16x2 + 9x2 (AC)2 = 25x2 AC = √25π‘₯2 AC = 5x Now, sin A = (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’ 𝐴)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ sin A = 𝐡𝐢/𝐴𝐢 sin A = 4π‘₯/5π‘₯ sin A = πŸ’/πŸ“ Similarly, cos A = (𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ 𝐴)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ cos A = 𝐴𝐡/𝐴𝐢 cos A = 3π‘₯/5π‘₯ cos A = πŸ‘/πŸ“ Given, 𝐭𝐚𝐧 𝐀=πŸ’/πŸ‘ cosec A = 1/sin⁑𝐴 = 1/(4/5) = πŸ“/πŸ’ sec A = 1/cos⁑〖 𝐴〗 = 1/((3/5) ) = πŸ“/πŸ‘ cot A = 1/tan⁑𝐴 = 1/(4/3) = πŸ‘/πŸ’

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