Trig functions (1 Viewer)

kpad5991 · Jan 28, 2018

Please help with 19b, c

https://imgur.com/gallery/ZWmLu

fan96 · Jan 28, 2018

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.

kpad5991 · Jan 28, 2018

Can you please please show the working im not getting it.

fan96 · Jan 28, 2018

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 $\begin{aligned} A_\text{sector} - A_\text{triangle} = 2\end{aligned}$

Trig functions (1 Viewer)

kpad5991

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fan96

617 pages

kpad5991

Member

fan96

617 pages

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