1. ## Trig functions

https://imgur.com/gallery/ZWmLu

2. ## Re: Trig functions

b) use formulae for area of sector and area of triangle.

c) repeat b) for the new area. By trying to solve for $\theta$ you should arrive at that equation naturally.

4. ## Re: Trig functions

b)

\begin{aligned} A_\text{sector}&= \frac{1}{2} r^2 \theta \\ &= \frac{\theta}{2} \end{aligned}

\begin{aligned} A_\text{triangle}&= \frac{1}{2} ab \sin C \\ &= \frac{1}{2}\sin{\theta} \end{aligned}

\begin{aligned} A_\text{sector} - A_\text{triangle} = 1 = \frac{\theta}{2} - \frac{1}{2} \sin \theta\end{aligned}

\begin{aligned} \frac{\theta}{2} - \frac{1}{2} \sin \theta &= 1 \\ \sin \theta &= \theta - 2\end{aligned}

Solve that using your answer from part a). If you don't want to photocopy the graph you can just put the equation into Wolfram Alpha and take the answer from that (remember to convert to degrees afterwards). As long as you understand the concept of using a graph to solve the equation it doesn't matter.

c) repeat part b) for \begin{aligned} A_\text{sector} - A_\text{triangle} = 2\end{aligned}

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