## Week 4 – Math 10

This week was short, and on top of that, I missed a day because I was at a field trip for my science honours class. But even though it was a short week, I still learned more about trigonometry, specifically word problems and how to use them to find angles and side lengths, something we’ve been learning over the whole unit.

To find the angle of a triangle, you can use two side lengths for the equation (sin/cos/tan) xº $= \frac {side1}{side2}$, for example: sin xº $= \frac {5}{9}$. With that equation (using the example for the following), you find x by isolating it as in xº $= sin^{-1} (\frac {5}{9})$, and then you will have the value of xº.

To find a side length, you would use one side tenth, and an angle for the equation (sin/cos/tan) xº $= \frac {n}{side length} or \frac {side length}{n}$, For example: sin 31º $\frac {n}{15}$. With that equation (using the example for the following), you find n by isolating it, in this case 15 $\cdot$ sin 31º $= n$. If it were sin 31º $= \frac {15}{n}$, you would use the equation $n \cdot$ sin 31º = 15 meaning you would have to divide both sin 31º (cancelling it out) and 15 to find n.

I made a simple word problem to find the side length of a triangle based on the height of a person and the suns angle to find the length of the persons shadow: