Can someone help me with these problems?
![Can someone help me with these problems class=](https://us-static.z-dn.net/files/db6/68dbeea2ec5966eaa3079e5a2f412ffb.png)
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:
see explanation
Step-by-step explanation:
16
Given
sinΘ = - [tex]\frac{\sqrt{3} }{2}[/tex] , find the related acute angle
Θ = [tex]sin^{-1}[/tex]([tex]\frac{\sqrt{3} }{2}[/tex]) = 60°
Since Θ is in quadrant 4, then
Θ = 360° - 60° = 300°
17
Given secΘ = - 2, then using cosx = [tex]\frac{1}{secx}[/tex] , then
cosΘ = [tex]\frac{1}{-2}[/tex] = - [tex]\frac{1}{2}[/tex], hence
Θ = [tex]cos^{-1}[/tex]( [tex]\frac{1}{2}[/tex] = 60° ← related acute angle
Since Θ is in quadrant 3 then
Θ = 180° + 60° = 240°