Given that tan(θ)= 3/2, find the other five trigonometric function of θ
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tan@ = opposite side / adjacent side
Given tan@ = 3 / 2
So , Opposite side = 3
Adjacent side = 2
By Pythagoras Theorum :- Hypotenuse^2 = Base^2 + Altitude^2
hyp^2 = 3^2 + 2^2
hyp^2 = 9 + 4
hyp^2 = 13
hypotenuse = √13
Sin@ = Opposite side / Hypotenuse = 3 / √13
Cos@ = Adjacent side / Hypotenuse = 2 / √13
Cot@ = Adjacent side / Opposite side = 2 / 3
Cosec@ = 1/Sin@ = √13 / 3
Sec@ = 1/Cos@ = √13 / 2
Given tan@ = 3 / 2
So , Opposite side = 3
Adjacent side = 2
By Pythagoras Theorum :- Hypotenuse^2 = Base^2 + Altitude^2
hyp^2 = 3^2 + 2^2
hyp^2 = 9 + 4
hyp^2 = 13
hypotenuse = √13
Sin@ = Opposite side / Hypotenuse = 3 / √13
Cos@ = Adjacent side / Hypotenuse = 2 / √13
Cot@ = Adjacent side / Opposite side = 2 / 3
Cosec@ = 1/Sin@ = √13 / 3
Sec@ = 1/Cos@ = √13 / 2
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Sketch a right triangle such that the sides are 3 (opposite θ) and 2 adjacent to θ.
The hypotenuse is sqrt(3^2 + 2^2) = sqrt(13)
Now use that right triangle to define the other five functions associated with θ.
If necessary, review those definitions in your textbook
It will be good practice for the next time, and there will be a next time!
The hypotenuse is sqrt(3^2 + 2^2) = sqrt(13)
Now use that right triangle to define the other five functions associated with θ.
If necessary, review those definitions in your textbook
It will be good practice for the next time, and there will be a next time!
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sin Ө = 3 / √13
cos Ө = 2 / √13
csc Ө = √13 / 3
sec Ө = √13 / 2
cot Ө = 2/3
cos Ө = 2 / √13
csc Ө = √13 / 3
sec Ө = √13 / 2
cot Ө = 2/3