if sinteta = 3/5 then cot theta equals
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cosø=4/5 . sorry for inveted photo
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Given
sin∅ = 3/5
We know that,
Sin∅ = P/H
=> P/H = 3/5
=> P : H = 3 : 5
So, let P be 3x and H be 5x
By Pythagoras theorem,
B (base) = √(H² - P²)
=> B = √{ (5x)² - (3x)²}
=> B = √(25x² - 9x²)
=> B = √(16x²)
=> B = 4x
So we know that,
cos∅ = B/H
=> cos∅ = 4x/5x
=> cos∅ = 4/5
Cot∅ = cos∅/sin∅
=> Cot∅ = 4/5 ÷ 3/5
=> Cot∅ = 4/5 × 5/3
=> Cot∅ = 4/3
Hope it helps dear friend ☺️✌️
sin∅ = 3/5
We know that,
Sin∅ = P/H
=> P/H = 3/5
=> P : H = 3 : 5
So, let P be 3x and H be 5x
By Pythagoras theorem,
B (base) = √(H² - P²)
=> B = √{ (5x)² - (3x)²}
=> B = √(25x² - 9x²)
=> B = √(16x²)
=> B = 4x
So we know that,
cos∅ = B/H
=> cos∅ = 4x/5x
=> cos∅ = 4/5
Cot∅ = cos∅/sin∅
=> Cot∅ = 4/5 ÷ 3/5
=> Cot∅ = 4/5 × 5/3
=> Cot∅ = 4/3
Hope it helps dear friend ☺️✌️
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