How Can Trigonometry Help Find an Obtuse Angle in a Circle Sector?

[Needhelp2]

A sector of a circle, let's say AOB with circle centre O and radius 5cm has a chord subtended from A to B. This chord forms a triangle with centre 0. Angle 0 isθradians, and the area of triangle A0B is 8cm². Given that angle AOB is obtuse, findθ.

I worked out Sin^-1(0.64)= 0.694, but this is not an obtuse angle and I don't know how to finish the problem (Sadface) any help would be greatly appreciated!

Thank you!

 

[Unknown008]

You never learned how to solve for angles? (Wondering)

Let's say sin(x) = 0.5

Then, the critical value of x is $\dfrac{\pi}{6}$

The values of x will be = $\dfrac{\pi}{6}$, $\pi - \dfrac{\pi}{6}$, $\dfrac{\pi}{6} + 2\pi$, $3\pi - \dfrac{\pi}{6}$, etc

For sine, the values are in the 1st and 2nd quadrant, for tan, 1st and 3rd quadrant, and for cos, 1st and 4th quadrant.

 

[CaptainBlack]

A sector of a circle, let's say AOB with circle centre O and radius 5cm has a chord subtended from A to B. This chord forms a triangle with centre 0. Angle 0 isθradians, and the area of triangle A0B is 8cm². Given that angle AOB is obtuse, findθ.

I worked out Sin^-1(0.64)= 0.694, but this is not an obtuse angle and I don't know how to finish the problem (Sadface) any help would be greatly appreciated!

Thank you!

You are looking for the solutions of \(\sin(\theta))=0.64\). If you sketch the \(\sin\) curve you will see that for \(\theta\) in the range \(0\) to \(2\pi\) you have a solution at about \(\theta=0.694\), and another at \(\theta=\pi-0.694\). The first of these is an acute angle (about 36.6 degrees) and the other obtuse.

CB

 

[Needhelp2]

thank you! I did know how to solve equations using sin and cos etc, but I didnt realize I could bring that knowledge to solve this problem!