Answer:
[tex]\text{A. }\frac{4\sqrt{14}}{{7}}[/tex]
Step-by-step explanation:
In any right triangle, the tangent of an angle is equal to its opposite side divided by its adjacent side. Therefore, we can form a right triangle with non-right angle [tex]\theta[/tex] and its opposite side [tex]\sqrt{7}[/tex] and its adjacent side [tex]5[/tex].
By definition, [tex]\csc \theta=\frac{1}{\sin\theta}[/tex].
In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle. To find the hypotenuse, use the Pythagorean Theorem: [tex]a^2+b^2=c^2[/tex], where [tex]c[/tex] is the hypotenuse, or longest side, of the right triangle and [tex]a[/tex] and [tex]b[/tex] are the two legs of the triangle.
Solving, we get:
[tex]5^2+\sqrt{7}^2=c^2,\\25+7=c^2,\\c^2=32,\\c=\sqrt{32}=4\sqrt{2}[/tex]
Therefore, we have:
[tex]\csc \theta = \frac{1}{\sin \theta}=\frac{1}{\frac{\sqrt{7}}{4\sqrt{2}}},\\\\\csc \theta=1\cdot \frac{4\sqrt{2}}{\sqrt{7}},\\\\\csc \theta =\frac{4\sqrt{2}}{\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\boxed{\text{A. }\frac{4\sqrt{14}}{{7}}}[/tex]
