April 3

# Week 7– Math 10 Trigonometry questions

1. In the picture we can see AC is 50, and we can find out angle ACB is 21˚, we need figure out the length of BC. The angle ACB is the reference angle, so the side AC is the HYP, the side BC is ADJ, so we need use cos ratio. Usually, we around the results to the nearest tenth.

The equation is cos 21˚= $(\frac{x}{50})$ the second step is: Remove X to the left side, so we could get a equation like the picture showing, x=50(cos21˚) then we got the result.

2. In this right triangle PCQ, the angle PCQ is 90 degrees, we already know the side PC is 23, and the side CQ is 40,  we need to figure out the angle CPQ. The angle y is the reference angle, so the side PC is the ADJ, the side CQ is the OPP, so we use tan ratio. And the results we need around to the nearest degrees.

The equation is tan y˚= $(\frac{40}{23})$ And the second step is leave the y alone, the equation must be y˚= $tan^-1\cdot (\frac{40}{23})$ so by the end, we could use calculator to get the result.

3. In this right triangle, the angle NML is 90˚, and the angle MNL is 30˚, we could easily figure out angle MLN is (90˚- 30˚=60˚), so b=60˚. The side ML is 27 and it is OPP side, the side NL is HYP side, so we need use sin ratio. Then the result is 54.

The equation is sin 30˚= $(\frac{27}{a})$ And the second step is to leave the a alone, so I moved the a to the left side and have the equation $(\frac{27}{sin30˚})$ And the last step is using calculator to calculate the results.