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An acute angle θ is in a right triangle with sin θ = 6/7 . What is the value of cot θ
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What have you tried till now?
Well,\[\tan \theta = {opposite \over adjacent}\]From \(\sin \theta\), we know that "opposite" is 6.
Also, you must must find the adjacent side through Pythagorean Theorem.\[6^2 + b^2 = 7^2\]
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Yeah, also recall that\[\cot \theta = {adjacent \over opposite}\]
so thats right? :)
\[b = \sqrt{7^2 - 6^2} = \sqrt{13}\]Yay!
ohh so 6/√13 ?
No, your answer is right.
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ok i gt confused for a seccc
love ur pic btw
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