Question

Solve each equation for the angle in standard position, for 0° ≤ 0 < 360° (nearest tenth, if necessary).

a) tan 0 = 1 / √3
b) 2cos 0= √3

Answer 1

Explanation:

a) To solve the equation tan θ = 1/√3, we can find the angle whose tangent is 1/√3 by taking the inverse tangent (arctan) of 1/√3.

θ = arctan(1/√3)

θ ≈ 30.0°

Therefore, the angle in standard position that satisfies tan θ = 1/√3 is approximately 30.0°.

b) To solve the equation 2cos θ = √3, we can isolate the cosine term by dividing both sides of the equation by 2.

cos θ = √3 / 2

Now, we can find the angle whose cosine is √3/2 by taking the inverse cosine (arccos) of √3/2.

θ = arccos(√3/2)

θ ≈ 30.0°

Therefore, the angle in standard position that satisfies 2cos θ = √3 is approximately 30.0°.