# Maths help

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#2

(Original post by

How do they know that triangle BCG is a right angled triangle?

**Student 999**)How do they know that triangle BCG is a right angled triangle?

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**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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#4

(Original post by

How do they know that triangle BCG is a right angled triangle?

**Student 999**)How do they know that triangle BCG is a right angled triangle?

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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#5

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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#6

(Original post by

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!

**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!

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#7

(Original post by

The point is, it doesn't matter if the triangles are similar or not. You can find what the question asks simply.

**mxllyhar**)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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#8

(Original post by

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.

**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.

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#9

(Original post by

Your clarification was great though. Do you study maths?

**mxllyhar**)Your clarification was great though. Do you study maths?

(Used to teach it.)

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#11

(Original post by

That's nice, did you like teaching? What age group did you teach?

**mxllyhar**)That's nice, did you like teaching? What age group did you teach?

But we are getting a bit off topic. Sorry, OP!

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#12

**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!

Proof of SAS

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#13

(Original post by

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

**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

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#14

(Original post by

Sorry my bad I skimmed it too quickly however the ratio between 40:20√5 and 20:8√5 are not the same.

**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.

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#16

(Original post by

How do they know that triangle BCG is a right angled triangle?

**Student 999**)How do they know that triangle BCG is a right angled triangle?

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#17

(Original post by

You mean 20√5:40 and 20:8√5 which is correct

**The A.G**)You mean 20√5:40 and 20:8√5 which is correct

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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#18

(Original post by

It isn’t. It doesn’t say that.

**spacegirl3**)It isn’t. It doesn’t say that.

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Thanks people,I’ve realized that I accidentally did the ratios the wrong way around

You can infer that it is a right angled triangle from when they used trigonometry.

(Original post by

It isn’t. It doesn’t say that.

**spacegirl3**)It isn’t. It doesn’t say that.

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#20

(Original post by

Well, it doesn't say that, but it is!

**Pangol**)Well, it doesn't say that, but it is!

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