Question

How do you find `tan theta`, given `cos theta = sqrt3 / 2` and `0° < theta < 90°`?

Answer 1

`tan t = sqrt3/3`

Explanation:

`cos t = sqrt3/2`. Find sin t
`sin^2 t = 1 - cos^2 t = 1 - 3/4 = 1/4`
`sin t = +- 1/2`
Because t is in Quadrant 1, therefor, sin t is positive
`sin t = 1/2`
`tan t = sin t/(cos t) = (1/2)/(sqrt3/2) = 1/sqrt3 = sqrt3/3`
Note. The trig table also gives --> `tan t = tan (pi/6) = sqrt3/3`

Answer 2

`tantheta=sqrt3/3`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(2/2)sin^2theta+cos^2theta=1`

`•color(white)(x)tantheta=sintheta/costheta`

`"given "costheta=sqrt3/2" and "0< theta < 90`

`theta" is in the first quadrant where sin/tan are positive"`

`sintheta=sqrt(1-cos^2theta)`

`color(white)(sintheta)=sqrt(1-(sqrt3/2)^2)=sqrt(1-3/4)=sqrt(1/4)`

`rArrsintheta=1/2`

`rArrtantheta=(1/2)/(sqrt3/2)`

`color(white)(tantheta)=1/2xx2/sqrt3=1/sqrt3xxsqrt3/sqrt3=sqrt3/3`