Question

An acute angle θ is in a right triangle with sin θ = 6/7 . What is the value of cot θ?

Answer 1

sin t = 6/7
cot t = cos t / sin t
cos t = √(1 - (6/7)²) = √( 1 - 36/49) = √(13/49) = √13/7
cot t = √13/7 / 6/7 = √13 / 6

Answer 2

`cot\theta=(√(13))/(6)`

Explanation:

Step 1

Find the value of the
`cos\theta`

we have that

`sin\theta=(6)/(7)`

Remember that

`sin^(2) \theta+cos^(2) \theta=1`

Substitute the value of
`sin\theta`

`(6)/(7)^(2)+cos^(2) \theta=1`

`(36)/(49)+cos^(2) \theta=1`

`cos^(2) \theta=1-(36)/(49)`

`cos^(2) \theta=(13)/(49)`

square root both sides

`cos \theta=(√(13))/(7)`

Step 2

Find the value of
`cot\theta`

we know that

`cot\theta=(cos\theta)/(sin\theta)`

substitute the values

`cot\theta=((√(13))/(7))/((6)/(7))=(√(13))/(6)`