You cannot add sines together like that. eg sin(90) = 1 but sin(180) is not 1 + 1. It's because the sine graph is a curve. And a sine can never be greater than 1.

So let's call J the midpoint of GI.

Then GJ = 20 and so sin(GHJ) = 20/25 and cos(GHJ) = 15/25

There is a formula for double angles:

Applying that here, we can say sin(GHI) = 2 x 20/25 x 15/25 = 96/100

Have a look here:

http://www.mathsisfun.com/algebra/trigo … index.html

Hope that helps.

Bob

]]>1. What is the sine of an acute angle whose cosine is 7/25?

4. In triangle GHI, we have GH = HI = 25 and GI = 30. What is \sin\angle GIH?

5. In triangle GHI, we have GH = HI = 25 and GI = 40. What is \sin\angle GHI? (Note: This is NOT the exact same as the previous problem!)

I'm working on these two problems. The triangle in question is not a right triangle. It's isosceles, but we can make some right triangles by adding the altitude from H to GI.

I can do number 4 -- my right triangle has sides KI= 15, HK= 20 and HI = 25. (I added K to be the foot of the altitude.) So \sin\angleGIH=20/25=4/5.

But I'm stuck on 5 -- \angle GHI is not part of my right triangle. Half of it is \angle GIk. I thought I could say \sin\angle GHI= \sin\angle GHk + \sin\angle IHk = 20/25 +20/25 = 8/5. But that is not the answer you guys got. Did you forget that GI is different in these two problems as well as the angle they are asking for? I'm really new to doing trig and not sure about my understanding of how things work yet, so I can't tell if it is me doing something wrong, or if it is just a misreading of the problem, or what. Thanks.

]]>Very good. Did you draw a diagram on that trig problem about the house?

]]>My bad! Thought it was GIH again. Your answer is correct for 5!

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