1. find sinθ if cosθ=1/2 and θ terminates in quadrant iv.
2. find cosθ if sinθ=(√2)/2 and θ terminates in quadrant i.
3. find tanθ if cosθ=-1/2 and θ terminates in quadrant ii.
4. find tanθ if sinθ=-1 and 0≤θ< 2π radians.

-(√3)/2, (√2)/2, -√3, and undefined

Step-by-step explanation:

There are two ways you can solve this.  One is with the Pythagorean identity:

sin²θ + cos²θ = 1

The other way is by knowing your unit circle.

1. From the unit circle, we know that cos θ = 1/2 at θ = π/3 and θ = 5π/3.  Since θ is in Quadrant IV, then θ = 5π/3.  sin (5π/3) = -(√3)/2.

We can check our answer using the Pythagorean identity:

sin²θ + cos²θ = 1

sin²θ + (1/2)² = 1

sin²θ + 1/4 = 1

sin²θ = 3/4

sin θ = ±(√3)/2

Since sine is negative in Quadrant IV, sin θ = -(√3)/2.

We can repeat these steps for the other questions.

2. sin θ = (√2)/2 at θ = π/4 and θ = 3π/4.  Since θ is in Quadrant I, θ = π/4.  Therefore, cos θ = (√2)/2.

3. cos θ = -1/2 at θ = 2π/3 and θ = 4π/3.  Since θ is in Quadrant II, θ = 2π/3.  Therefore, sin θ = (√3)/2, and tan θ = sin θ / cos θ = -√3.

4. sin θ = -1 at θ = 3π/2.  Therefore, cos θ = 0.  tan θ = sin θ / cos θ, so tan θ is undefined.

75

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.