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Let θ Be an Angle in Standard Position sinθ=49,tanθ>0\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quad

Question 67

Multiple Choice

Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
- sinθ=49,tanθ>0\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quad Find secθ\sec \theta


A) 96565- \frac { 9 \sqrt { 65 } } { 65 }
B) 659- \frac { \sqrt { 65 } } { 9 }
C) 94\frac { \sqrt { 9 } } { 4 }
D) 46565- \frac { 4 \sqrt { 65 } } { 65 }

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