Question

An acute angle θ is in a right triangle with sin θ = five sixths. What is the value of cot θ?
Answers:
five divided by the square root of eleven six divided by the square root of eleven square root of eleven divided by five square root of eleven divided by six

Answer 1

@phi

Answer 2

Can you explain more to me please

Answer 3

Do you know about pythagorean theorem?

Answer 4

Yes

Answer 5

can you write down the formula for this triangle ?
fill in the letters with numbers ?

Answer 6

a^2+b^2=c^2

Answer 7

yes, but put in the numbers (we know 2 out of the 3 sides, right ?)

Answer 8

So 5^2+6^2=c^2

Answer 9

almost. a and b are the legs. we only know one of the legs
c is the hypotenuse. we know the hypotenuse, right ?

Answer 10

Oh so it's 5^2+b^2=6^2?

Answer 11

yes. it does not matter which leg (a or b) we use)
next, expand 5^2 and 6^2

Answer 12

25+b^2=36

Answer 13

now add -25 to both sides, and then simplify

Answer 14

b^2=11

Answer 15

finally, "take the square root" of both sides
the left side becomes b

Answer 16

That part is the part I'm confused on

Answer 17

we get
\[ b= \sqrt{11} \]
we usually would leave it like that, unless we have to do calculations

Answer 18

Ohhhh okay so then there has to be more to it though, right?

Answer 19

square root "undoes" the square
i.e.
\[ \sqrt{b^2} = b\]
so you have to remember that.

Answer 20

so
\[ b^2 = 11\]
do the square root on both sides
\[ \sqrt{b^2} =\sqrt{11} \]
and then we remember square root undoes the square:
\[ b= \sqrt{11}\]

Answer 21

But that's not an answer choice

Answer 22

@phi

Answer 23

we now know *all 3 sides* of the right triangle