Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 Re: Help Me ! » trig problems » 2016-03-14 12:19:03
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.
#2 Re: Help Me ! » Trigonometry! » 2016-03-14 12:01:05
hi evene,
Q3. That equation can be re-written with only cos.
Now use
So now the equation has only cos(A). Subtract 1 from both sides so you can square to eliminate the square root sign.
Re-arrange and solve the quadratic. You will find two values for cos(A). One does not fit the original equation so can be discarded; the other does fit as you can easily show.
Bob
I'm working on this problem and having trouble with the last part--solving the quadratic. I'm ending up with a really strange looking quadratic, having to complete the square. Is that supposed to happen? Or am I making a silly math mistake that I'm just not catching? Thanks.
#3 Re: Help Me ! » Geometry Homework » 2016-02-09 06:18:44
hi numbergeek
Welcome to the forum.
You cannot do 260 - 260 x 0.25 because it doesn't take account of the bottom square properly. It isn't scaled down. Stick to working out the sloping area first and you should be OK.
Hope that helps,
Bob
Thanks, Bob. That helped me a lot. I hadn't thought about it that way and couldn't figure out what I was doing wrong.
#4 Re: Help Me ! » Geometry Homework » 2016-02-08 03:40:00
I am also working on this problem and I don't understand why 220 is not correct. The total surface area of the pyramid is 260. The scale factor is .25 (.5 squared). I have 260 - 260(.25) + surface area of the smaller square. Length of the larger square is 10, so smaller square side length is 5. So I think the correct calculation is 260 - 65 + 25 which is 220. but that answer is wrong.
Pages: 1