Question

1) a. Sketch a right triangle that has acute angle 0, with tan 0b. Find the exact values of the other five trigonometric ratios of 0. Rationalize denominators.the angle theta in degrees rounded to one

Answer 1

a)

b) The other 5 trigonometric ratios are:

`\begin{gathered} \sin\theta=(3√(58))/(58) \\ \\ \\ \\ \end{gathered}`
`\cos\theta=(7√(58))/(58)`
`\csc\theta=(√(58))/(3)`
`\sec\theta=(√(58))/(7)`
`\ctg\theta=(7)/(3)`

Step-by-step explanation:

The in a right triangle of the form:

the tan ratio is:

`\tan\theta=(opposite)/(adjacent)`

Then, we know:

`\tan\theta=(3)/(7)`

Then,

opposite = 7

adjacent = 3

We can draw:

For b, we need to know the formulas for the other 5 trigonometric ratios.

The formulas are:

`\begin{gathered} \sin(\theta)=(opposite)/(hypotenuse) \\ . \\ \cos(\theta)=(adjacent)/(hypotenuse) \\ . \\ \csc(\theta)=(1)/(\sin(\theta)) \\ . \\ \sec(\theta)=(1)/(\cos(\theta)) \\ . \\ \ctg(\theta)=(1)/(\tan(\theta)) \end{gathered}`

We can see that we need to know the length of the hypotenuse. We can use the pythagoren theorem to find it.

The pytagorean theorem tell us:

`Hypotenuse^2=opposite^2+adjacent^2`

Then:

`Hypotenuse^2=3^2+7^2`

And solve:

`\begin{gathered} Hypotenuse^2=9+49=58 \\ Hypotenuse=√(58) \end{gathered}`

Now we can find:

`\sin(\theta)=(oppos\imaginaryI te)/(hypotenuse)=(3)/(√(58))`

We need to radicalize the denominator. Thus: