Please view the attachment.

My working out

Note my answer is wrong. I assumed that the red and blue triangle (shown in the second attachment) are similar shapes.

I got 20 for the long side via Pythagoras's theorem.

16/12 = 1.333

Therefore splitting 20 in the ratio of 1:1.333 = 8.6:11.4.

a=11.4, b=8.6 --> this is incorrect

Please explain how to get the correct answer. Why is it wrong to do 16/12 to get a multiplier of 1.333?

My working out

Note my answer is wrong. I assumed that the red and blue triangle (shown in the second attachment) are similar shapes.

I got 20 for the long side via Pythagoras's theorem.

16/12 = 1.333

Therefore splitting 20 in the ratio of 1:1.333 = 8.6:11.4.

a=11.4, b=8.6 --> this is incorrect

Please explain how to get the correct answer. Why is it wrong to do 16/12 to get a multiplier of 1.333?

(edited 10 months ago)

All three triangles are similar (the two smaller ones and the larger one). 20 is the larger hypotenuse. Not sure why you think 16/12 is useful. For both the red and blue triangles, you need to relate the unknown a or b to the hypotenuse (of the red or blue triangle), so the useful side ratios for the larger triangle would be

relevant leg / hypotenuse

Tbh, you can interpret the relevant side ratios in terms of simple trig if you find that easier. Your 16/12 is tan(), whereas the releavant trig terms for the red and blue trianges are sin() and cos(). Sometimes its worth drawing the three triangles as separate ones, with the same orientation, so the correct side ratios are obvious

relevant leg / hypotenuse

Tbh, you can interpret the relevant side ratios in terms of simple trig if you find that easier. Your 16/12 is tan(), whereas the releavant trig terms for the red and blue trianges are sin() and cos(). Sometimes its worth drawing the three triangles as separate ones, with the same orientation, so the correct side ratios are obvious

(edited 10 months ago)

Original post by mqb2766

All three triangles are similar (the two smaller ones and the larger one). 20 is the larger hypotenuse. Not sure why you think 16/12 is useful. For both the red and blue triangles, you need to relate the unknown a or b to the hypotenuse (of the red or blue triangle), so the useful side ratios for the larger triangle would be

relevant leg / hypotenuse

Tbh, you can interpret the relevant side ratios in terms of simple trig if you find that easier. Your 16/12 is tan(), whereas the releavant trig terms for the red and blue trianges are sin() and cos()

relevant leg / hypotenuse

Tbh, you can interpret the relevant side ratios in terms of simple trig if you find that easier. Your 16/12 is tan(), whereas the releavant trig terms for the red and blue trianges are sin() and cos()

Not sure what you mean by " so the useful side ratios for the larger triangle would be relevant leg / hypotenuse"?

What is a relevant leg?

(Just trying to process what these bits mean)

I assumed I could 16 and 12 are both hypotenuse's for the smaller triangles so therefore using the ratio of their size I could split 20 accordingly to get a and b.

(edited 10 months ago)

Original post by As.1997

What do you mean by " so the useful side ratios for the larger triangle would be relevant leg / hypotenuse"?

What is a relevant leg?

What is a relevant leg?

I was editing as you replied. Try sketching the 3 similar triangles as separate ones with the same orientation (so the same angles occur in the same place). That way, the matching of the sides should be clear. You should see a and b correspond to the 16 and 12 in the larger triangle, so the 16 or 12 is the relevant leg ( similarity relationship for the red or blue triangle).

(edited 10 months ago)

Original post by mqb2766

I was editing as you replied. Try sketching the 3 similar triangles as separate ones with the same orientation (so the same angles occur in the same place). That way, the matching of the sides should be clear. You should see a and b correspond to the 16 and 12 in the larger triangle, so the 16 or 12 is the relevant leg ( similarity relationship for the red or blue triangle).

Is this the right idea?

Original post by As.1997

Is this the right idea?

Agree with the sketches, though Id make the difference between the shorter and longer leg obvious (blue triangle). If youre unsure, mark the top angle as theta, then you want for the red and bue triangles

cos(theta) = a/16 = ...

sin(theta) = b/12 = ...

and the ... can simply be read off the large black triangle (mark the 20 hypotenuse on) as the appropriate side ratio. Though as you note, once you have one you can do 20-... to get the second.

(edited 10 months ago)

Original post by mqb2766

Agree with the sketches, though Id make the difference between the shorter and longer leg obvious (blue triangle). If youre unsure, mark the top angle as theta, then you want for the red and bue triangles

cos(theta) = a/16 = ...

sin(theta) = b/12 = ...

and the ... can simply be read off the large black triangle as the appropriate side ratio.

cos(theta) = a/16 = ...

sin(theta) = b/12 = ...

and the ... can simply be read off the large black triangle as the appropriate side ratio.

Am I right to assume the a's I have marked are not actually equal to one another?

Original post by As.1997

Am I right to assume the a's I have marked are not actually equal to one another?

Definately. Theyre complementary (sum to 90) and are equal to the two non right angles of the large black triangle (which are not equal for a 3-4-5 triangle).

(edited 10 months ago)

Original post by mqb2766

Agree with the sketches, though Id make the difference between the shorter and longer leg obvious (blue triangle). If youre unsure, mark the top angle as theta, then you want for the red and bue triangles

cos(theta) = a/16 = ...

sin(theta) = b/12 = ...

and the ... can simply be read off the large black triangle (mark the 20 hypotenuse on) as the appropriate side ratio. Though as you note, once you have one you can do 20-... to get the second.

cos(theta) = a/16 = ...

sin(theta) = b/12 = ...

and the ... can simply be read off the large black triangle (mark the 20 hypotenuse on) as the appropriate side ratio. Though as you note, once you have one you can do 20-... to get the second.

I couldn't quite follow this. So the sketches are okay but the working is wrong?

I just added an attachment to show you what I understood

(edited 10 months ago)

Original post by As.1997

I couldn't quite follow this. So the sketches are okay but the working is wrong?

Your sketches are ok, but Id make the blue one a bit taller to emphasise that the height is the longer leg.

In the sketches, mark theta as top angle in all three triangles (theyre similar). Then for b (you can check your a working)

sin(theta) = opp/hyp = b//12 = 12/20

b/12 is the blue triangle, 12/20 is the black triangle. So b = ... Then do similar for a (or subtact from 20 to check)

Similar triangles (side ratios) and trig are fairly synonymous. Im only suggesting mixing them here to make clear how to match the triangles. A decent sketch of the triangles should make which side ratios are appropriate to equate without saying sin() = ...

(edited 10 months ago)

Original post by mqb2766

Your sketches are ok, but Id make the blue one a bit taller to emphasise that the height is the longer leg.

In the sketches, mark theta as top angle in all three triangles (theyre similar). Then for b (you can check your a working)

sin(theta) = opp/hyp = b//12 = 12/20

b/12 is the blue triangle, 12/20 is the black triangle. So b = ... Then do similar for a (or subtact from 20 to check)

Similar triangles (side ratios) and trig are fairly synonymous. Im only suggesting mixing them here to make clear how to match the triangles. A decent sketch of the triangles should make which side ratios are appropriate to equate without saying sin() = ...

In the sketches, mark theta as top angle in all three triangles (theyre similar). Then for b (you can check your a working)

sin(theta) = opp/hyp = b//12 = 12/20

b/12 is the blue triangle, 12/20 is the black triangle. So b = ... Then do similar for a (or subtact from 20 to check)

Similar triangles (side ratios) and trig are fairly synonymous. Im only suggesting mixing them here to make clear how to match the triangles. A decent sketch of the triangles should make which side ratios are appropriate to equate without saying sin() = ...

Is this what you meant?

I have just added another attachment, in the second attachment I have a concern regarding theta. As theta seems to be the same for the black and red triangle but it is something else for the blue triangle. How can we be sure that theta is the same for the black red and blue triangle?

(edited 10 months ago)

Original post by As.1997

Is this what you meant?

Yes. Id have done the a calculation for practice and used teh sum to 20 to check.

In a sense there are two ways to view similarity,

1) match one side in the relevant triangles and get a scale factor then apply that to the unknown side. So for b the scale factor (black to blue) would be based on the hypotenuse so 12/20 and then apply that to the black base to get b (blue base) so 12*12/20 = ...

2) Write down explicitly the side ratio in each triangle so b/12 = 12/20. ( = sin(theta) here).

Done correctly, both are fine, and correctly is just about making sure you argue about the right sides hence the sketch advice. But if youre unsure Id probably recommend the second way as you can explain (check) it in terms of simple trig. Though the first way is more common if you want to do area radio in terms of side ratio (or similar) so you square the relevant scale factor.

(edited 10 months ago)

Original post by mqb2766

Yes. Id have done the a calculation for practice and used teh sum to 20 to check.

In a sense there are two ways to view similarity,

1) match one side in the relevant triangles and get a scale factor then apply that to the unknown side. So for b the scale factor (black to blue) would be based on the hypotenuse so 12/20 and then apply that to the black base to get b (blue base) so 12*12/20 = ...

2) Write down explicitly the side ratio in each triangle so b/12 = 12/20. ( = sin(theta) here).

Done correctly, both fine, and correctly is just about making sure you argue about the right sides hence the sketch advice. But if youre unsure Id probably recommend the second way as you can explain (check) it in terms of simple trig.

In a sense there are two ways to view similarity,

1) match one side in the relevant triangles and get a scale factor then apply that to the unknown side. So for b the scale factor (black to blue) would be based on the hypotenuse so 12/20 and then apply that to the black base to get b (blue base) so 12*12/20 = ...

2) Write down explicitly the side ratio in each triangle so b/12 = 12/20. ( = sin(theta) here).

Done correctly, both fine, and correctly is just about making sure you argue about the right sides hence the sketch advice. But if youre unsure Id probably recommend the second way as you can explain (check) it in terms of simple trig.

"I'd have done the a calculation for practice and used the sum to 20 to check." - I think this is one way to check my doubts about whether the theta in the blue triangle was the same as in black and red triangles. If it wasn't then the side (a) would not have been 12.8 to make the 20.

(edited 10 months ago)

Original post by As.1997

Is this what you meant?

I have just added another attachment, in the second attachment I have a concern regarding theta. As theta seems to be the same for the black and red triangle but it is something else for the blue triangle. How can we be sure that theta is the same for the black red and blue triangle?

I have just added another attachment, in the second attachment I have a concern regarding theta. As theta seems to be the same for the black and red triangle but it is something else for the blue triangle. How can we be sure that theta is the same for the black red and blue triangle?

Your second diagram is correct and you used this at the start as you stated the triangles were simillar. However, you need to be sure why. If the angle in the bottom right is gamma (so common to both the black and blue triangles), for the large black triangle theta and gamma are complementary so

gamma + theta = 90

Therefore the top angle in the blue triamge must be theta as the other two are gamma and 90.

Really you should just note that all three triangles are right triangles, so as long as one of the other angles match theta (red/black) and gamma (blue/black) they must be similar as as long as two angles match, the third must as well (sum to 180).

(edited 10 months ago)

Original post by mqb2766

Your second diagram is correct and you used this at the start as you stated the triangles were simillar. However, you need to be sure why. If the angle in the bottom right is gamma (so common to both the black and blue triangles), for the large black triangle theta and gamma are complementary so

gamma + theta = 90

Therefore the top angle in the blue triamge must be theta as the other two are gamma and 90.

Really you should just note that all three triangles are right triangles, so as long as one of the other angles match theta (red/black) and gamma (blue/black) they must be similar as as long as two angles match, the third must as well (sum to 180).

gamma + theta = 90

Therefore the top angle in the blue triamge must be theta as the other two are gamma and 90.

Really you should just note that all three triangles are right triangles, so as long as one of the other angles match theta (red/black) and gamma (blue/black) they must be similar as as long as two angles match, the third must as well (sum to 180).

I think I got you - I highlighted it in the attachment below *.

Thanks a lot for your help!

(edited 10 months ago)

Original post by As.1997

I think I got you - I highlighted it in the attachment below *.

Thanks a lot for your help!

Thanks a lot for your help!

Pretty much spot on, though youd be sent to the back of the line for mixing greek and latin "names" for the angles. Its just offensive.

Original post by mqb2766

Pretty much spot on, though youd be sent to the back of the line for mixing greek and latin "names" for the angles. Its just offensive.

Oh oh : O

Original post by As.1997

Oh oh : O

Just "for interest", when you drop an altitude (the extra perpendicular line in the original black triangle) in a right triangle, you always get two other similar triangles (red and blue). It can be used to prove pythagoras using a similar, similar argument and for non right angled triangles dropping an altitude is a common thing to do when you prove the sin and cos rules, and 1/2 related to thales circle theorem proof (thinking how the right angle is split using a radius rather than a perpendicular).

(edited 10 months ago)

Original post by mqb2766

Just "for interest", when you drop an altitude (the extra perpendicular line in the original black triangle) in a right triangle, you always get two other similar triangles (red and blue). It can be used to prove pythagoras using a similar, similar argument and for non right angled triangles dropping an altitude is a common thing to do when you prove the sin and cos rules, and 1/2 related to thales circle theorem (thinking how the right angle is split using a radius rather than a perpendicular).

Thanks!

Original post by As.1997

Thanks!

Could be worth noting (unless my working are wrong) that you don't have to use similar shapes. I think that you could draw out the 3 triangles, find what a + b is, find what a^2 - b^2 is and use the difference of two squares to solve.

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