In ΔHIJ, the measure of ∠J=90°, the measure of ∠I=72°, and JH = 1.2 feet. Find the length of HI to the nearest tenth of a foot.

1.3

Step-by-step explanation:

Looking at the triangle, the first thing we can notice is that it is a right triangle, so we can use trigonometric ratios to solve for missing sides.

The side the problem is asking us to find is leg HI, which happens to be the hypotenuse of triangle HIJ.

We also know one side length: HJ= 1.2 ft, and one angle measure: ∠I=72°.

In these types of problems, there are two trigonometric ratios we can look to:

Sine or  sin(x)= opposite/hypotenuseCosecant or csc(x)= hypotenuse/opposite

You can actually use either of these to solve your problem, but for convenience let's use cosecant.

Cosecant of 72° will be equal to HI (the hypotenuse length) divided by HJ (the length of the opposite leg of angle I) Cosecant of 72° will be equal to HI divided by 1.2 (substitute the given length for HJ)1.05146222424 (plug csc(72°) into a calculator)=HI/1.2 1.05146222424*1.2=HI (multiply each side of the equation by 1.2) HI=1.26175466909 (simplify) HI≈1.3 (round the simplified number to the nearest tenth, as the problem requests)

disclaimer: i'm not at all qualified to be teaching trigonometry haha (but i hope this helps you!!)

please please please let me know if i made any mistakes

Are you talking about question 7 eleven?