# Tutor profile: Dominique C.

## Questions

### Subject: Trigonometry

Say you have a triangle with angles of 30 degrees, 60 degrees, and 90 degrees. The side opposite from the 60 degree angle is equal to 9 inches. What are the lengths of the other two sides?

This triangle is an identity. We refer to the 30-60-90 and 45-45-90 triangles as Special Triangles. The ratio of the angles: $$30:60:90$$ correspond to the side lengths: $$1:\sqrt{3}:2$$. From the property of similar triangles, we know that any triangle with angles of 30, 60, and 90 degrees will have side lengths that are multiples of the ratio above. This fact allows us to rewrite the ratio as: $$1x:x\sqrt{3}:2x$$ where $x$ is the value we need to multiply by to get the similar triangle. The problem stated that the side length measuring 9 inches is opposite of the 60 degree angle. This tells us that we can rewrite our ratio of a similar triangle as the following: $$1x:9:2x$$. This means that $$9 = x\sqrt{3}$$. Let's solve for $$x$$. $$ x\sqrt{3} = 9$$ $$ x = \frac{9}{\sqrt{3}}$$ $$ x = \frac{9\sqrt{3}}{3}$$ $$ x = 3\sqrt{3}$$ Let's plug in $x$ into our ratio of the similar triangle. $$1( 3\sqrt{3}):9:2( 3\sqrt{3})$$ $$3\sqrt{3}:9:6\sqrt{3}$$ So the sides $$3\sqrt{3}:9:6\sqrt{3}$$ (all with units of inches) correspond to the angles $$30:60:90$$ (with units of degrees).

### Subject: Geometry

Find the volume of a cube with side lengths of 5 cm.

The formula for the volume of a cube is the area of the base times the height. $$ V = B \times h $$ The base of a cube is a square, which has has an area of length times width. $$ V = l \times w \times h $$ All of the sides of the cube are of equal length, so we know the following: $$ w = l, h = l $$ Therefore we can rewrite the formula as: $$ V = l \times l \times l = l^{3} $$ We know that $$ l = 5 cm $$ from the given value in the problem, so we substitute this value into our equation. Remember, the cube affects the value, $5$, and the units $cm$. $$ V = (5cm)^{3} $$ $$ V = 125cm^{3} $$

### Subject: Algebra

Solve the following system of equations for the variables x and y: $$ 3x + 4y = 9 $$ $$ x - 4y = 11 $$

We can see that in the first equation, the coefficient of y is 4. In the second equation, the coefficient of y is -4. $$ 3x + 4y = 9 $$ $$ x - 4y = 11 $$ We can see that by adding these terms, they would cancel out. So, we add the first and second equations. $$ (3x + x) + (4y - 4y) = (9 + 11) $$ Let's simplify: $$ 4x = 20 $$ Now we have one equation with one unknown variable. Solving for x we find: $$ x = 5 $$ Now we just need to plug $$x = 5$$ into one of the original equations above to solve for y $$ 3x + 4y = 9 $$ $$ 3(5) + 4y = 9 $$ $$ 15 + 4y = 9 $$ $$ 4y = 9 - 15 $$ $$ 4y = -6 $$ $$ y = \frac{-6}{4} $$ $$ y = \frac{-3}{2} $$ Therefore, $$ x = 5, y = \frac{-3}{2} $$

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