# Thread: Stuck on simple trig

1. ## Stuck on simple trig

Hi everyone

I'm stuck on (ii).. I can't figure it out. I breezed through the rest of the questions but I haven't got this yet. It needs to be done by Monday so I will work on it during that time and if I don't get it I will check back here. If anyone wants to help, can you try and give me just a hint rather than the full explanation? I would love to work it out "myself". Haha

2. ## Re: Stuck on simple trig

I'm not sure, but it seems that if you would at b from 2 different points you might get something:

b^2 = a^2 - d^2 and b^2 = c^2 - d^2 , hope this helps,

dokrbb

3. ## Re: Stuck on simple trig

Hmm.. Thanks for the reply, but those are not right angles necessarily so Pythagoras' theorem won't work. Just confirming, I know the answer to question (i), but not (ii). I think (ii) might have something to do with constructing other triangles... Maybe somehow joining the two that are on JF?

4. ## Re: Stuck on simple trig

oh, sorry - I briefly looked at the image and stupidly saw there a right angle,

No worries

6. ## Re: Stuck on simple trig

I think this might be the key!

7. ## Re: Stuck on simple trig

Okay so I don't know if anyone is actually reading this but I know I can now solve the question if I prove that angle IHJ = angle HJI + angle CDI. I am struggling to prove this but I feel like I have made progress.

8. ## Re: Stuck on simple trig

To be franc i am not getting an idea. In case you show your work on the problem i might get some idea.

9. ## Re: Stuck on simple trig

Are you hinting that you know the answer? Look I have done as much as I can to this point

10. ## Re: Stuck on simple trig

I might be wrong again, but maybe you should use the congruence properties for (Side Angle Side) for the triangles HGF and CGB...

I would really want to see the final answer

11. ## Re: Stuck on simple trig

Okay well I figured it out. In the picture where I cut the triangles out and put them together, it is a right angled triangle. You can say the large triangle has internal angles IHJ, GHF and the other angle is the sum of the four angles HJI, CDI, CBG, and GFH. Now, you could say angle IHJ has an angle x and so GHF has an angle of 90 - x (because it's a straight angle in the original diagram, subtracting a right angle from the square, therefore the two angles add to 90). Now, since those are two angles of the large triangle, we can find the third (y) by saying
y = 180 - x - (90 - x)
y = 180 - x - 90 + x
y = 90
Therefore, the large triangle created from the four small triangles has a right angle at the vertex that all triangles meet at. This triangle has a hypotenuse 4d and a median b (splits the hypotenuse into two lengths of 2d). Now, there is a theorem that says that the median of a right angled triangle that meets the hypotenuse is half the length of the hypotenuse (basically because if you split the triangle along the median it creates two isosceles triangles, here is a great proof Lesson Median drawn to the hypotenuse of a right triangle). This means that b = (4d)/2, and b = 2d. So there you go

12. ## Re: Stuck on simple trig

cool, thanks for posting