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# Tutor profile: Joanna H.

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Joanna H.
Doctorate in Mathematics and 5 years of tutoring experience
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## Questions

### Subject:Trigonometry

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Question:

Prove that $$\sin(60) = \frac{\sqrt{3}}{2}$$

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Joanna H.

For this proof we can use an equilateral triangle which has all sides equal to 1. Additionally for equilateral triangles, since all the sides are the same length, all the angles are also all the same. We also know that for all triangles the total of all the angles must equal 180 degrees. Therefore each of the angles in an equilateral triangle are 60 degrees (180 divided by 3). Next, draw a line from the middle of one of the corners to the opposite edge of the triangle so that it makes a right angle with the edge and cuts the edge exactly in half. This effectively divides the triangle into two smaller triangles. Now we focus on one of these smaller triangles. Since they are right angled triangles we can use the well known trigonometric identity (often referred to as SOHCAHTOA) which says that $$sin(x) = \frac{opposite}{hypotenuse}, \quad cos(x) = \frac{adjacent}{hypotenuse}, \quad \tan{x} = \frac{opposite}{adjacent}$$ where $$x$$ is an angles of the triangle other than the 90 degree angle. Opposite and adjacent refer to the length of the side opposite or adjacent to the angle and hypotenuse is the length of the hypotenuse. Therefore in our case looking at one of smaller triangles we have that $$sin(60) = \frac{opposite}{hypotenuse}$$. We know that the $$hypotenuse = 1$$ because we specified at the beginning that the original triangle had sides of length 1. To find the length of the opposite side we can use the Pythagorean theorem which says that for any right angled triangle $$a^2 +b^2 =c^2$$ where c is the length of the hypotenuse and a and b are the lengths of the other 2 sides of the triangle. In our case we can let a = the length opposite the $$60^o$$ angle which is what we still need to find, and b = the length adjacent to the $$60^o$$ angle which we know is $$\frac{1}{2}$$. Therefore $$a^2+\frac{1}{4} = 1$$ which can be rearranged to give $$a^2 = \frac{3}{4}$$ Taking the square root gives $$a = \frac{\sqrt{3}}{2}$$ Therefore if the length opposite to the $$60^o$$ angle is $$\frac{\sqrt{3}}{2}$$ then we have that $$sin(60) = \frac{\sqrt{3}}{2}$$. QED (A diagram would greatly aid this explanation)

### Subject:Basic Math

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Question:

Which number comes next in this sequence $$2,3,5,7,11,13...$$

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Joanna H.

You should have recognized this sequence as the list of prime numbers. Prime numbers are numbers which have exactly 2 factors: 1 and itself. Therefor the next number in this sequence is $$17$$. This is because all the numbers in between ($$14, 15, 16$$) have more than 2 factors.

### Subject:Algebra

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Question:

Find $$z$$ in the following equation $$3(2z-3)+6 = 21$$

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Joanna H.

Step 1) Minus 6 from both sides: $$3(2z - 3) = 15$$ Step 2) Divide both sides by 3: $$2z - 3 = 5$$ Step 3) Add 3 to both sides: $$2z = 8$$ Step 4) Divide both sides by 2: $$z = 4$$

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