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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 (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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.