I need some help with this...
I tried a few way but they did not work, that was for solving a different problem
Look at angle x. Side 19 is the hypotenuse of the triangle. For angle x, is side 13 the opposite leg of the adjacent leg?
so how do i set it up to solve
I usually know how to solve it its just i don't know how to set it up
When you are solving right triangles using trig, the sides of the triangle have to be named. The side opposite the right angle is always the longest side of the triangle and is always named the hypotenuse. Look at the figure below. You see that the hypotenuse is always the side opposite the right angle? |dw:1436137160349:dw|
Ooh okay i see
Good. A right triangle also has two other sides. The other two sides are the sides that form the right angle. They are both called legs. See figure below. You see which sides are the legs? |dw:1436137437906:dw|
In trig of right triangles, the sine, cosine, ad tangent are ratios of the lengths of the sides of a right triangle. The hypotenuse has its own name and is the only side with that name. The two legs are called legs, but that is not good enough because we need to distinguish between them. We have to have a different name for each leg.
When we deal with these problems, there is always one acute angle we are considering. For each of the acute angles, we can name the legs more specifically. One leg is opposite the acute angle, and the other leg is adjacent the acute angle.
The legs are named as shown in the figure with respect to angle x. The leg next to angle x is the adjacent leg. The other leg is the opposite leg. |dw:1436138535481:dw|
Here is your problem. |dw:1436138676748:dw|
You are looking for the measure of angle x. You are given the lengths of the opposite leg and the hypotenuse. The trig function that relates the opposite leg to the hypotenuse is the sine function. \(\sin \theta = \dfrac{opp}{hyp} \) \(\sin x = \dfrac{13}{19} \) To find x, first divide 13 by 19, then find the inverse sine of that number.
Join our real-time social learning platform and learn together with your friends!