Student 999
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How do they know that triangle BCG is a right angled triangle?
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mxllyhar
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(Original post by Student 999)
How do they know that triangle BCG is a right angled triangle?
Maybe you could post the full solution and I could look. Nothing suggests the triangle is a right-angle triangle.
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Student 999
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(Original post by mxllyhar)
Maybe you could post the full solution and I could look. Nothing suggests the triangle is a right-angle triangle.
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Pangol
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(Original post by Student 999)
How do they know that triangle BCG is a right angled triangle?
It is easy to work out the length CF.

You can then use the given ratio of lengths to find the lengh CG.

In triangle CDF, you now know the lengths of the two sides around the labelled angle.

In triangle BCG, you also now know the lengths of the two sides around the labelled angle.

But these angles are the same, so you should be able to show that these two triangles are the similar. Triangle CDF is right-angled, so therefore triangle BCG is.

I can't help feeling that there should be an easier way of showing these triangles are similar, but this will do!
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mxllyhar
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Yeah, they haven't assumed that the triangle is right angle. They know FDC is right angle hence they used Pythagoras's Theorem to find the length of FC. They then used the fact 3CG = 2GF and then used the alternate angle property to find the angle marked in the question. Then simply plugged the values into the area of a triangle formula.
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mxllyhar
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(Original post by Pangol)
It is easy to work out the length CF.

You can then use the given ratio of lengths to find the lengh CG.

In triangle CDF, you now know the lengths of the two sides around the labelled angle.

In triangle BCG, you also now know the lengths of the two sides around the labelled angle.

But these angles are the same, so you should be able to show that these two triangles are the similar. Triangle CDF is right-angled, so therefore triangle BCG is.

I can't help feeling that there should be an easier way of showing these triangles are similar, but this will do!
The point is, it doesn't matter if the triangles are similar or not. You can find what the question asks simply.
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Pangol
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(Original post by mxllyhar)
The point is, it doesn't matter if the triangles are similar or not. You can find what the question asks simply.
Yeah, I can see that you don't need this to answer the question. I was answering the question asked by the OP. I hadn't seen the solution by that point and was assuming they used this as part of their working.
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mxllyhar
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(Original post by Pangol)
Yeah, I can see that you don't need this to answer the question. I was answering the question asked by the OP. I hadn't seen the solution by that point and was assuming they used this as part of their working.
Your clarification was great though. Do you study maths?
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Pangol
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(Original post by mxllyhar)
Your clarification was great though. Do you study maths?
Insofar as one never stops learning, yes!

(Used to teach it.)
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mxllyhar
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(Original post by Pangol)
Insofar as one never stops learning, yes!

(Used to teach it.)
That's nice, did you like teaching? What age group did you teach?
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Pangol
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(Original post by mxllyhar)
That's nice, did you like teaching? What age group did you teach?
Apart from a very few special cases, I only taught A Level - Maths, Further Maths, Physics, and Computing on one occasion. I adored it.

But we are getting a bit off topic. Sorry, OP!
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The A.G
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(Original post by Pangol)
It is easy to work out the length CF.

You can then use the given ratio of lengths to find the lengh CG.

In triangle CDF, you now know the lengths of the two sides around the labelled angle.

In triangle BCG, you also now know the lengths of the two sides around the labelled angle.

But these angles are the same, so you should be able to show that these two triangles are the similar. Triangle CDF is right-angled, so therefore triangle BCG is.

I can't help feeling that there should be an easier way of showing these triangles are similar, but this will do!
The sides found in both triangles are not corresponding/complementary thus you can’t use them to prove similarly however OP can assume that triangle BCG is a right angle then use Pythagoras theory to find length BG, if the ratio between two pairs of complementary triangle sides are the same then both triangles would be similar.

Proof of SAS
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Pangol
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(Original post by The A.G)
The sides found in both triangles are not corresponding/complementary thus you can’t use them to prove similarly however OP can assume that triangle BCG is a right angle then use Pythagoras theory to find length BG, if the ratio between two pairs of complementary triangle sides are the same then both triangles would be similar.

Proof of SAS
Really? The sides I'm talking about are the adjacent and hypotenuse of the respective triangles. In CFD, they have lengths 20 and 20root5. In BCG, they have length 8root5 and 40. The ratios of the two side lengths are the same in both cases, root5, and considering they are either side of angles we know to be the same, one triangle is a scaled-up version of the other. (Yes, I know there's a reflection as well.)
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Pangol
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(Original post by The A.G)
Sorry my bad I skimmed it too quickly however the ratio between 40:20√5 and 20:8√5 are not the same.
But the ratios you want to look at are 20√5:20 and 40:8√5.
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The A.G
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You mean 20√5:40 and 20:8√5 which is correct
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Buddingdentist08
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(Original post by Student 999)
How do they know that triangle BCG is a right angled triangle?
It isn’t. It doesn’t say that.
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Pangol
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(Original post by The A.G)
You mean 20√5:40 and 20:8√5 which is correct
That is one way of doing it, but I know what I mean.

If you had two triangles where the sides adjacent to angles you know are the same are 1 and 2 in one triangle and 4 and 8 in the other, then the fact that 2/1 = 8/4 shows that one is a scalled up version of the other.
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Pangol
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(Original post by spacegirl3)
It isn’t. It doesn’t say that.
Well, it doesn't say that, but it is!
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Student 999
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#19
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Thanks people,I’ve realized that I accidentally did the ratios the wrong way around

(Original post by spacegirl3)
It isn’t. It doesn’t say that.
You can infer that it is a right angled triangle from when they used trigonometry.
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Buddingdentist08
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(Original post by Pangol)
Well, it doesn't say that, but it is!
Sorry. I was looking at the wrong triangle lol
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