In right triangles, SOHCAHTOA tells us that , and we know that  and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

 Use a calculator or reference to approximate cosine.

 Isolate the variable term.

Thus, .

In right triangles, SOHCAHTOA tells us that , and we know that  and leg. Therefore, a simple substitution and some algebra gives us our answer.

 Use a calculator or reference to approximate cosine.

 Isolate the variable term. 

Thus, .

The question asks for you to find the perimeter of the given figure. The figure has twelve sides total, of two varying lengths. One length is given to you, 4. The other length must be solved for using either the sine or tangent functions. However, one can arrive to answer more quickly by recognizing that the drawn triangle is actually a 3-4-5 triangle, where 3, 4, and 5 corresponds to each of the sides of the triangle. This is a pythagorean triple and this ratio should be easily remembered.

Thus if 3 is the missing side, and there are eight sides of length 3 and four sides of length 4, one can arrive to the answer:

You can draw out your scenario like a triangle:

Now, you know that this means:

Using your calculator, you can utilize the inverse  function to calculate the degree measure of the angle:

This rounds to  degrees.

To solve this question, you must know SOHCAHTOA. This acronym can be broken into three parts to solve for the sine, cosine, and tangent.

We can use this information to solve our identity.

Dividing by a fraction is the same as multiplying by its reciprocal. 

The sine divided by cosine is the tangent of the angle.

The cotangent of the angle of a triangle is the adjacent side over the opposite side. The answer is 

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, . 

This is . Rounding, this is .  Since  is in the third quadrant, your value must be positive, as the tangent function is positive in that quadrant.

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, , or . Since  is in the third quadrant, your value must be positive, as the tangent function is positive in this quadrant.

Based on our information, we can draw this little triangle:

Since we know that the tangent of an angle is , we can say:

This can be solved using your calculator:

 or 

Now, to solve for , use the Pythagorean Theorem, , where  and  are the legs of a triangle and  is the triangle's hypotenuse. Here, , so we can substitute that in and rearrange the equation to solve for :

Substituting in the known values:

, or approximately . Rounding, this is .

You can begin by imagining a little triangle in the second quadrant to find your reference angle. It would look like this:

The tangent of an angle is:

For our data, this is:

Now, since this is in the second quadrant, the value is negative, given the periodic nature of the tangent function.