Tutor profile: Samantha G.

Evaluate $$cot(\frac{\pi}{6})$$

There are a few things we need to know for this question. 1. The trig identity for cotangent is the reciprocal of tangent. Since $$tangent=\frac{sine}{cosine}$$, then $$cotangent=\frac{cosine}{sine}$$ because we flip the fractions. 2. You need the coordinate point for $$\frac{\pi}{6}$$ which is $$(\frac{\sqrt{3}}{2}. \frac{1}{2})$$. It is important to remember the points on the unit circle are read as $$(cos, sin)$$. They follow alphabetical order, which is a quick way to remember. 3. Now we are ready to solve! $$cotangent=\frac{cosine}{sine}$$ $$cotangent\frac{\pi}{6}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$ There is a 2 on the numerator and denominator, so they cancel $$cotangent\frac{\pi}{6}=\frac{\sqrt{3}}{1}$$ $$cotangent\frac{\pi}{6}=\sqrt{3}$$

Assume that $$\angle ABD$$ and $$\angle DBC$$ are complementary. If $$m\angle ABD$$ is twice $$m\angle DBC$$. What is the $$m\angle ABD$$?

First of all, we need to know what complementary means. Complementary angles add to $$90^{\circ}$$. Therefore we can say that $$\angle ABD + \angle DBC = 90$$. Next, we need to substitute in our values. If $$m\angle ABD$$ is twice $$m\angle DBC$$, then $$m\angle ABD$$ is two times larger than $$m\angle DBC$$. Lets say that $$\angle DBC = x$$, then $$\angle ABD =2x$$. Now we plug those values into our equation and solve for x. $$\angle ABD + \angle DBC = 90$$ $$2x + x = 90$$ $$3x = 90$$ $$\frac{3x}{3} = \frac{90}{3}$$ $$x = 30$$ We defined our variables as $$\angle DBC = x$$ and $$\angle ABD =2x$$. Therefore, $$m\angle DBC = 30^{\circ}$$ and $$\angle ABD =2*30=60^{\circ}$$

Solve for x in the given equation: $$ 2x+5=9 $$

$$ 2x+5=9 $$ to solve for x, you first need to combine like terms. $$ -5 $$ $$ -5 $$ the opposite of addition is subtraction, so we need to subtract 5 from both sides $$2x=4 $$ simply $$ \frac{2x}{2}= \frac{4}{2} $$ the opposite of multiplication is division, so we divide both sides by 2 $$x=2$$ simplify and you get your answer