# Tutor profile: Nicholas R.

## Questions

### Subject: Trigonometry

A) What is the hypotenuse of a right triangle that has two shorter sides measuring 4 inches and 10 inches respectively. B) what are the angles of the acute corners of the right triangle?

For part A to solve the hypotenuse of a right triangle we will apply the Pythagorean Theorem which states that the square of the hypotenuse is the summation of the squares of the two other sides or a^2 + b^2 = c^2. Here c is the hypotenuse and a and b are the two other sides. So plugging in the values we know the resulting formula is 4^2 + 10^2 = c^2. Simplifying we arrive at sqrt( 16 +100 )= c. So c = 2 * sqrt (29) which is approximately 10.8 inches. For part be to determine the angles we must utilize the trigonometric functions sine, cosine or tangent. Any of the functions can be used but you must first know how to apply it. A simple way to remember what each functions ratio is equal to is to remember the term SOH-CAH-TOA. In this case SOH stands for the sine of some angle = Opposite/ Hypotenuse. CAH stands for the cosine of some angle = Adjacent / Hypotenuse and TOA stands for tangent of some angle = Opposite / Adjacent. We will use SOH for this example. To find the angle occurring at the corner of 4 and the hypotenuse 2*sqrt(29) we will say sine(theta) = (10/(2*sqrt(29)). Theta being the unknown angle. To solve for theta we will isolate it by conducting the inverse sine (sin^-1) to both sides. Thus we arrive at theta = sin^-1(10/(2*sqrt(29)). Simplifying this formula we get the answer for the angle theta to be approximately 68.2 degrees. To solve for the other angle we will use the rule that for any right triangle the three angles add up to 180 degrees. Since one angle we know is 90 degrees and the second we solved for is 68.2 degrees we will subtract both from 180. Thus the third angle is 180-90-68.2 = 21.8 degrees.

### Subject: Mechanical Engineering

There is an 8 meter beam attached rigidly to a wall horizontally. A weight of 70 newtons is applied to the non-fixed end of the beam. Another force of 30 newtons is applied upward 3 meters away from the fixed end. What is the resulting moment at the fixed end of the beam (point A)?

To determine the moment about the fixed end we must determine the summation of moments across the beam about the fixed end. A moment is the measure of a forces tendency to cause rotation about a specific point or axis. The formula for a moment (M) with force (F) is M=F*x with x being the distance from the point or axis the force is imposed on. To calculate the resulting moment on the fixed end we will determine the moments created by both the weigh and upward forces. We will also assume the direction of the resulting moment is going to be clockwise which will be represented in our summation equation as positive. So, summation of moments about the fixed point (A) is M(A)= -30*3 + 70*8. Simplifying the formula and solving for M(A) = 470 N*m. Since this value is positive and we assumed clockwise to be positive then the resulting moment at the fixed point A is clockwise. Finally the resulting moment at fixed point A is 470 N*m clockwise.

### Subject: Calculus

Find the area under the curve for y=(x^2)+5 from x=4 to x=7.

In order to determine the area under the curve we will utilize integration. Next, let's integrate the curve with the formula y=(x^2)+5. In order to do so we will integrate the variable x^2 and the constant 5 separately. When integrating a variable the variable will be raised exponentially by 1 and then divided by the new resulting number of the exponent. So for x^2, the integral will be (X^(2+1))/(2+1). The result is (X^3)/3. For a constant, we will imagine it is multiplied by a variable raised to the zeroth power. So for 5, we will say 5*x^0. Using the rule integrating variables the new value becomes 5*(x^(0+1))/1. Thus, the equation becomes 5*x. Finally, when two values are added together in an integration formula the values will be integrated by the respective rules and then combined after integration. Thus the integral of y=(x^2)+5 becomes (x^3)/3 + 5*x. Since we are finding the area from x=4 to x=7 you must subtract the values of 7 plugged into the integrated formula minus the value of 4 plugged into the integrated formula. Thus substituting both values into the formula we arrive at 149.33 - 41.33 = 108. Thus the area under the curve of y=(x^2)+5 from 4 to 7 is 108.

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