If cot theta = √9-x² , find sec theta and tan theta
Answers
Step-by-step explanation:
Hypotenuse : BC = 13
Adjacent Side : AB = 13 (Adjacent to ∠ Θ)
Opposite Side : AC = ? (Opposite to ∠ Θ)
To find the Opposite Side - AC , we use Pythagoras theorem ((Hypotenuse)2 = (Leg 1)2 + (Leg 2)2)
____________
AC = √ BC2 - AB2
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AC = √ 132 - 112
____________
AC = √ 169 - 121
__
AC = √ 48
Now, we will simplify square root
_________________
AC = √ 2 x 2 x 2 x 2 x 3
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AC = 2 + 2 √ 3 (pairs taken out)
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AC = 4 √ 3
So Cosec, Sec and Cot will be as follows -
Cosec Θ = Hypotenuse / Opposite Side
__
Sin Θ = 13 / 4√ 3
Cot Θ = Adjacent Side / Opposite Side
__
Tan Θ = 11 / 4√ 3
Sec Θ = Hypotenuse / Adjacent Side
Sec Θ = 13/11
Step-by-step explanation:
Hypotenuse : BC = 13
Adjacent Side : AB = 13 (Adjacent to ∠ Θ)
Opposite Side : AC = ? (Opposite to ∠ Θ)
To find the Opposite Side - AC , we use Pythagoras theorem ((Hypotenuse)2 = (Leg 1)2 + (Leg 2)2)
____________
AC = √ BC2 - AB2
____________
AC = √ 132 - 112
____________
AC = √ 169 - 121
__
AC = √ 48
Now, we will simplify square root
_________________
AC = √ 2 x 2 x 2 x 2 x 3
__
AC = 2 + 2 √ 3 (pairs taken out)
__
AC = 4 √ 3
So Cosec, Sec and Cot will be as follows -
Cosec Θ = Hypotenuse / Opposite Side
__
Sin Θ = 13 / 4√ 3
Cot Θ = Adjacent Side / Opposite Side
__
Tan Θ = 11 / 4√ 3
Sec Θ = Hypotenuse / Adjacent Side
Sec Θ = 13/11