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posledela
3 years ago
9

Find cotθ if θ terminates in Quadrant III and secθ = - 2.

Mathematics
2 answers:
sukhopar [10]3 years ago
7 0
sec(x) =  \frac{1}{cos(x)}  \\  \\ -2 =  \frac{1}{cos(x)}  \\  \\ cos(x) =  -\frac{1}{2}

If you know your sine/cosine values you know that when:
 cos(x) = 1/2 , sin(x) = √(3)/2
The fact that theta, or "x" as I've been using... terminates in Quadrant III means both cosine and sine will be negative and then cotangent will be positive.

cot(x) = cos(x) / sin(x)
cot(x) = (-1/2) / (-√(3)/2)
cot(x) = (1/2) * (2/√(3))

cot(x) = 1/√(3)
or rationalized as √(3)/3




pochemuha3 years ago
5 0
O.2 is the answer too yout qeustion sir.
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c). Area of ABCD = Area of ΔABD + Area of ΔBCD

  Area of ΔABD = AD×BD×Sin(\frac{24.36}{2})

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