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!

 

[CellCoree]

I can provide some guidance on how to approach this problem. Firstly, let's review the applications of trigonometry in this scenario. Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. In this case, we have a sector of a circle, which can be thought of as a triangle with one curved side. The chord subtended from A to B creates an additional triangle within the sector. Trigonometry can be used to calculate the angles and sides of these triangles.

Now, let's address the specific problem at hand. We are given the area of triangle A0B and the radius of the circle, which can help us find the length of the chord. We can use the formula for the area of a triangle, A = 1/2 * base * height, to find the length of the chord. In this case, the base is the length of the chord, and the height is the distance from the center of the circle to the chord. Using the given area of 8cm2 and the radius of 5cm, we can solve for the length of the chord to be 8cm.

Next, we can use the Law of Cosines to find the angle AOB. The Law of Cosines states that for a triangle with sides a, b, and c and opposite angles A, B, and C, the following equation holds true: c2 = a2 + b2 - 2ab * cos(C). In our case, we know the lengths of sides a and b (5cm and 8cm, respectively) and we can solve for the angle AOB to be approximately 1.23 radians.

Finally, we can use the Law of Sines to find the angle θ. The Law of Sines states that for a triangle with sides a, b, and c and opposite angles A, B, and C, the following equation holds true: sin(A)/a = sin(B)/b = sin(C)/c. We know the lengths of sides a and b, and we just solved for the angle AOB. Therefore, we can plug in these values to solve for the angle θ to be approximately 0.694 radians.

It is important to note that the angle θ may have multiple solutions, as the Law of Sines allows for a triangle to have two possible angles with the same sine value. However, based on the