Question

Can anyone help? Please and thank you!

Answer 1

Answer:

$csc(\theta)=\frac{6}{5}\\\\sin(\theta)=\frac{6\sqrt{11}}{11}\\\\cot(\theta)=\frac{6\sqrt{11}}{11}$

Step-by-step explanation:

This problem asks that one to finds the value of some basic trigonometric functions. Remember, the trigonometric ratios in a right triangle are the following,

$sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}$

One also has the reciprocal functions for trigonometric ratios these are as follows,

$csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{adjcanet}{opposite}\\\\cot(\theta)=\frac{adjacent}{opposite}$

This problem gives on the information that the side adjacent to the angle ($\theta$) has a measure of (5) units. The hypotenuse, side opposite the right angle measures (6) units.

One can use the Pythagorean theorem to solve for the unknown side. The Pythagorean theorem states the following,

$a^2+b^2=c^2$

Where (a) and (b) are the sides adjacent to the right angle of a right triangle. (c) is the side opposite the right angle; the hypotenuse. Substiute the given values in and solve for the unknown,

$(5)^2+(b)^2=(6)^2\\25+b^2=36\\b^2=11\\b=\sqrt{11}$

The problem asks one to find certain ratios. To do this, simplify substitute the sides into their respective positions in the ratio.

$csc(\theta)=\frac{hypotenuse}{opposte}=\frac{6}{5}$

$sin(\theta)=\frac{hypotenuse}{opposite}=\frac{6}{\sqrt{11}}=\frac{6\sqrt{11}}{11}$

$cot(\theta)=\frac{adjacent}{opposite}=\frac{5}{\sqrt{11}}=\frac{5\sqrt{11}}{11}$