Question

Helllllllllllllllllllllllllllp please

Answer 1

`3 \tan {}^(2) \theta - \tan\theta= 0 \\ \tan\theta(3 \tan \theta- 1) = 0✓\\\\\tan\theta = 0 \\\boxed{\theta = 0°} ✓\\\\ 3 \tan\theta - 1 = 0 \\3 \tan \theta = 1 \\ \tan \theta = (1)/(3) \\ \theta = { \tan }^( - 1) ( (1)/(3) ) \\\boxed{\theta = 18.4°}✓`

★ Value of the ∅ is 0° and 18.4°.

Answer 2

0° and 18°

Explanation:

3tan²β - tanβ = 0

tanβ(3tanβ - 1) = 0

tan β = 0 ⇒
`\beta _(1)` = 0°

3tan β - 1 = 0 ⇒ tan β =
`(1)/(3)` ⇒
`\beta _(2)` = arctan (
`(1)/(3)` ) ≈ 18°