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Trigonometry Applications of trigonometry!

Needhelp

New member
Apr 9, 2012
17
A sector of a circle, lets 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 dont know how to finish the problem (Sadface) any help would be greatly appreciated!!

Thank you!
 

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Unknown008

Member
Feb 1, 2012
20
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

Well-known member
Jan 26, 2012
890
A sector of a circle, lets 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 dont 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
 
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Needhelp

New member
Apr 9, 2012
17
thank you! I did know how to solve equations using sin and cos etc, but I didnt realise I could bring that knowledge to solve this problem!