Question

Can someone help me with these problems?

Answer 1

Answer:

The answer to your question is:

a) 240 and 300°

b) 120 and 240°

Step-by-step explanation:

You can see the right triangle below (sorry for my handwriting)

I suggest you to memorize the right triangles for 30°, 60° and 45°, they will be very useful.

To solve these questions, you need to work with the right triangle of 60°.

a)

Remember that  sine Ф = opposite side / hyp

             then      sine 60 = √3 / 2   the answer is 60 but it is negative so you need to look for these angles in thtethird and forth quadrangles.

third q = 180 + 60 = 240°    forth q = 360 - 60 = 300°

b)

Work with the same triangle

                   sec Ф = hyp / adjacent side

                   sec 60 = 2 / 1 = 2         the answer is 60°

but as it is negative you have to look for angles in the second and third quadrangle.

They will be 180 -60 = 120°  and 180 + 60 = 240°

Answer 2

Answer:

see explanation

Step-by-step explanation:

16

Given

sinΘ = - $\frac{\sqrt{3} }{2}$ , find the related acute angle

Θ = $sin^{-1}$($\frac{\sqrt{3} }{2}$) = 60°

Since Θ is in quadrant 4, then

Θ = 360° - 60° = 300°

17

Given secΘ = - 2, then using cosx = $\frac{1}{secx}$ , then

cosΘ = $\frac{1}{-2}$ = - $\frac{1}{2}$, hence

Θ = $cos^{-1}$( $\frac{1}{2}$ = 60° ← related acute angle

Since Θ is in quadrant 3 then

Θ = 180° + 60° = 240°