Question

An acute angle θ is in a right triangle with sin θ = two thirds . What is the value of cot θ?

Answer 1

check the picture below.

Answer 2

Answer:

six divided by the square root of thirteen

Step-by-step explanation:

hey there,

< sin θ = $\frac{O}{H}$

So that means O = 2 and H = 3. In order to find cot θ, first let's find tan θ.

tan θ = $\frac{O}{B}$

We only know what O is equal to, not B. So let's draw out a triangle.

7  

◢   6

 B

As you can see (sorry for the poor triangle), this is a right triangle. In order to find an unknown part, use $a^2 + b^2 = c^2$!

$6^2 + B^2 = 7^2$

B = ±√13

Obviously, a side of a triangle can't be negative, so it stays positive. Now we can find tangent!

tanθ =  $\frac{6}{\sqrt{13} }$

But, we're not done here. We're trying to find cotθ.

cotθ = $\frac{1}{tan}$θ

$\frac{1}{\frac{6}{\sqrt{13} } }$ = $\frac{\sqrt{13} }{6}$

That's your final answer! >

Hope this helped! Feel free to ask anything else.