Question

help

Answer 1

What's your question ?

Answer 2

^

Answer 3

@dude

Answer 4

cot(xº)=\(\large{\frac{1}{tan(xº)}}\)

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
cot(xº)=\(\large{\frac{1}{tan(xº)}}\)
\(\color{#0cbb34}{\text{End of Quote}}\)
Was not the answer to the first question x'D

Answer 6

@kaylak you need to use the formula that @dude gave you.

`x` is the degree.

\(\large\bf{cot~ 60^{o} = \frac{1}{tan~ 60^{o}}}\)

We then look at this:


We take note that \(\bf{tan~ 60^{o} = \sqrt{3}}\)

So we that....

\(\Large\bf{cot ~60^{o} = \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}}\)

Answer 7

We draw out the next problem,

Answer 8

We follow that...



Since we know of the opposite and the hypotenuse we would follow by: `Sine`

\(\large\bf{sin(x)=\frac{opposite}{hypotenuse}}\)

\(\Large\bf{sin(x)=\frac{\color{red}{15}}{\color{red}{30}}}\)

We inverse this.

\(\Large\bf{sin^{-1}(\color{Red}{\frac{15}{30}})=x}\)

What is `x`?

Answer 9

For 3, look at the conversion table for part 1.

Answer 10

b?

Answer 11

Is that one for 2 or 3?

Answer 12

1 b 2 b 3 c 4 b or c?

Answer 13

?

Answer 14

1 and 2 are correct.

Answer 15

@Vocaloid im a bit confused on 3 honestly.

Answer 16

#3 is based on the unit circle which you just have to memorize

Answer 17

so 3 is c what's 4

Answer 18

oh okie :3

Answer 19

this is a special type of triangle called a 30-60-90 triangle

the side across from 30 is "x" = 4
the side across from 60 is xsqrt(3)

so the missing side is 4sqrt(3)

Answer 20

so it is b yay i do know this stuff i try