# Tutor profile: Tanner L.

## Questions

### Subject: Trigonometry

A right triangle has an angle angle of 23deg and a side length of 5.2m adjacent to it. Find the hypotenuse of the triangle.

This problem requires the use of the inverse cosine trigonometric function. We know that the cosine function can be written as such: $$ \cos{\theta} = \frac{adjacent}{hypotenuse} $$ Plugging in what we know, we get the following. For the purposes of this problem, we'll refer to the hypotenuse length as $$ H $$: $$ \cos(23\deg) = \frac{5m}{H} $$ Rearranging the above equation to isolate $$ H $$, we get the following: $$ H = \frac{5m}{\cos(23\deg)} $$ Finally solving the above equations gives: $$ H = 5.43m $$

### Subject: Mechanical Engineering

Using a 2D x and y plane as a frame of reference, there are 3 forces that all act at the origin (0,0). One force of 50N acts at $$60^o$$ from the positive x-axis and another force of 30N acts at $$120^o$$ from the positive x-axis. Find the magnitude and direction of the third in order to maintain equilibrium.

The best way to approach this problem is to break the given forces into x and y vectors, then summing the forces in the x and y direction to find the required vectors to maintain equilibrium. First, lets break the given forces into vectors: Force 1 x-direction: $$(50N)\cos{60} = 25N $$ y-direction: $$ (50N)\sin{60} = 43.3N $$ Force 2 x-direction: $$(30N)\cos{120} = -15N $$ y-direction: $$ (30N)\sin{120} = 25.9N $$ Now summing the forces in the x and y direction gives: $$ \sum{Fx +25N -15N} $$ $$ Fx = 15N -25N $$ $$ Fx = -10N $$ $$ \sum{Fy +43.3N + 25.9N} $$ $$ Fy = -43.3N - 25.9N $$ $$ -69.2N $$ Now we just need to convert these x and y components into a magnitude with a direction: $$ \vert F \vert = \sqrt{(-10N^2) + (-69.2N^2)}$$ $$ \vert F \vert $$ = 69.92N $$ \arctan({\frac{-69.2N}{-10N})} = 81.77\deg + 180\deg = 261.78\deg$$ *The reason for adding 180deg to our final angle is because our x and y components of our force vector show the vector acting in the third quadrant of the plane, therefore 81.77deg would not make sense.

### Subject: Calculus

Find the following indefinite integral: $$ \int\mathrm{6x }^{5} - 18x^2 + 7\mathrm{d}x $$

To make solving this integral easier, we can look at this as three separate integrals by breaking up each of the individual terms like so: $$ \int\mathrm{6x }^{5}\mathrm{d}x - \int_0^\infty \mathrm18x^2\mathrm{d}x + \int_0^\infty7\mathrm{d}x$$ Analyzing the first integral, we simply raise the power by one and divide by the new value like so: $$ \int\mathrm{6x }^{5}\mathrm{d}x = \frac{6x^6}{6} = x^6 $$ Analyzing the second integral using the same method: $$ -\int\mathrm{18x }^{2}\mathrm{d}x = \frac{-18x^3}{3} = -6x^3 $$ The third term is an integer value with no variable, so we can simply add an $$x$$ to this term like so: $$ \int\mathrm{7} \mathrm{d}x = 7x $$ Now, because this is an indefinite integral, we must also add a constant variable to the end of our solution to represent the possibility of a constant in our general solution. With this in mind, the final solution becomes: $$ x^6 - 6x^3 + 7x + c $$

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