Using the formula for area of a triangle equal to , drawing  and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae , which can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

Therefore, the area of the triangle is 95.38 square centimeters.

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 4, 8, and 9 given in this problem. Heron's formula can be described as:

 where R is the area of the triangle and s is the semiperimeter equivalent to . Using these formulas and plugging in, we get:

Therefore the area of  is 15.998 meters.

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 1.1, 3.7, and 4.3 given in this problem. Heron's formula can be described as:

 where R is the area of the triangle and s is the semiperimeter equivalent to . Using these formulas and plugging in, we get:

Therefore the area of  is 1.83 feet.

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

The squaring of the sine in the denominator makes the sine term positive, i.e.

The numerator is the double angle formula for sine:

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

The final reduced equation is:

Thus proving that the equivalence is false.

The correct answer is . Rather than memorizing all three Pythagorean Relationships, you can memorize only , then simply divide all terms by  to get the formula that relates  and . Alternatively, you can divide all terms of  by  to get the formula that relates  and . The former is demonstrated below.

The correct answer is . Rather than memorizing all three Pythagorean Relationships, you can memorize only , then simply divide all terms by  to get the formula that relates  and . Alternatively, you can divide all terms of  by  to get the formula that relates   and . The former is demonstrated below.

Getting every term in an expression in terms of sine and cosine functions is a popular way to verify trigonometric identities or complete trigonometry proofs. These two trig functions are more commonly used over their counterparts secant, cosecant, tangent, and cotangent. Moreover, getting all terms of an expression in strictly sine and cosine may help you to spot and then substitute , or it may help you spot other functions that can be reduced or simplified. Other general techniques to aid in verifying trigonometric identities are:

1. Know the eight basic relationship and recognize alternative forms of each. These are: , , , , , , , and .
2. Understand how to add and subtract fractions, reduce fractions, and transform fractions into equivalent fractions
3. Know how to factor, and know special product techniques (i.e. difference of squares)
4. Only work on one side of the equation at a time
5. Choose the side of the equation that looks more complicated and attempt to transform it into the other side of the equation
6. Alternatively, you may transform each side of the equation into the same form. If doing this, you may need to do scratch work, then go back and nicely organize the two transformed sides into one clean looking verification.
7. Avoid substitutions that introduce radicals.
8. Multiply the numerator and denominator of a fraction by the conjugate of either.
9. Simplify a square root of a fraction by using conjugates to transform it into the quotient of a perfect squares. 

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods. 

The correct answer to how to best begin this verification is "Simplify a square root of a fraction by using conjugates to transform it into the quotient of perfect squares." This technique will be effective because while the left side of the equation has a square root and the right hand side does not, eliminating that square root will be crucial to verifying the identity. The verification of this identity is demonstrated below:

Let's also explore why the incorrect answers are incorrect. 

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions, while failing to get rid of the square root function. Next, this problem does not lend itself to substituting a Pythagorean Identity. Pythagorean Identities are most helpful when you can use them to replace something like  with . They can also be helpful if you want to replace something like  with  (while this seems more complicated to replace one term with two terms, if all of the rest of the terms in the expression were all cosines, this could be helpful.) Next, consider factoring. While often a helpful strategy, this method lends itself best to when you have polynomial terms such as , which could be factored into . This problem does not have this type of expression, so this isn't helpful to us. Finally, reducing fractions can be a great strategy when there are fractions to reduce. Do not fall into the trap of thinking that you are able to cancel the  or  expressions on the left hand side of the original equations. If you wanted to split this up, the only way to properly do it would be to get . Notice how large this expression is, and notice that it is getting more complicated rather than simpler and closer to . Often times, when trying to verify trig identities, you may start with one approach and find it not fruitful, and need to go back to the start and try again. Always remember that there may be another way to approach the problem; be flexible and be willing to experiment and try more than one approach!

This question is asking us to fill in the first four blanks by adding the appropriate trigonometric function that satisfies each. By defining an angle  on the coordinate plane, and labelling the distances x, y, and r, we can ask ourselves which trig functions relate each of the following sides together. Many students use the acronym SOHCAHTOA to remember these. Using the diagram below, we can see that , , and . Knowing each of these three answers, the fourth answer (based on the available answer choices) must be .

Next, we need to fill in the two question marks. To do this, simply substitute the info from the first part of the proof to get that  and .

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods. 

The correct answer to how to best begin this verification is "Get everything in terms of sines and cosines." This technique will be effective because left side presently does not have any like terms and it is an easy way to create like terms, which may then combine, reduce, cancel each other out, or simplify. The verification of this identity is demonstrated below:

Let's also explore why the incorrect answers are incorrect. 

In order to substitute a Pythagorean Identity, we usually want to see either , , or . None of these are present in our expression, so this is a non-starter. Next, let's consider the answer choice: "Multiply the numerator and denominator of a fraction by the conjugate of each other." Conjugates are applicable when the numerator or denominator are comprised of two or more terms. In our expression, each numerator and denominator consists of one term only, so this strategy isn't pertinent to this verification. Likewise, there are not any terms present that we can factor, and there are not any terms present that we can reduce.