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trigonometry problem

I attached the question. How can I solve this?

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Original post by thebrahmabull
I attached the question. How can I solve this?


Hint: The triangles are similar.
Original post by BuryMathsTutor
Hint: The triangles are similar.


Nice question
Original post by BuryMathsTutor
Hint: The triangles are similar.


thanks!
Reply 4
What is the answer?
Is the answer 36.87?

I'm in year 9 btw so don't say anything if i got it wrong....
Original post by Dizgurl
What is the answer?
Is the answer 36.87?

I'm in year 9 btw so don't say anything if i got it wrong....


Yeah, well I got the same answer too. 36.9, always round the answer to 3 significant figures. Nevertheless, same answer.

Ratio between 2 hypotenuses, 2M to 4M = 1:2.
So, BC and CD's length's will be split up in the ratio 1:2 aswell. BC = 1.2m and CD = 2.6m.

SOHCAHTOA
Angle we're finding = BAC
Opposite side = BC
Hypotenuse = AC

sin-1 (1.2 / 2) = 36.86989765
= 36.9 (3 s.f)
Reply 6
Original post by Chittesh14
Yeah, well I got the same answer too. 36.9, always round the answer to 3 significant figures. Nevertheless, same answer.

Ratio between 2 hypotenuses, 2M to 4M = 1:2.
So, BC and CD's length's will be split up in the ratio 1:2 aswell. BC = 1.2m and CD = 2.6m.

SOHCAHTOA
Angle we're finding = BAC
Opposite side = BC
Hypotenuse = AC

sin-1 (1.2 / 2) = 36.86989765
= 36.9 (3 s.f)


Cool, I just used the sine rule and similar triangles.
Original post by Dizgurl
Cool, I just used the sine rule and similar triangles.


Oh right lol. Yeah, the sine rule was possible but it's SOHCAHTOA is more preferred as we are dealing with Right-Angled Triangles. Nevertheless, nice!
Reply 8
Original post by Chittesh14
Oh right lol. Yeah, the sine rule was possible but it's SOHCAHTOA is more preferred as we are dealing with Right-Angled Triangles. Nevertheless, nice!


Thanks!
Ok, Ill use SOHCAHTOA next time....:smile:
Original post by Chittesh14


Ratio between 2 hypotenuses, 2M to 4M = 1:2.
So, BC and CD's length's will be split up in the ratio 1:2 aswell. BC = 1.2m and CD = 2.6m


Even assuming that you mean 2.4 instead of 2.6

I do not agree with this premise
Original post by Dizgurl
Thanks!
Ok, Ill use SOHCAHTOA next time....:smile:


It is fine to use sine and cosine rules even in right angled triangles

Sadly you do not have the correct answer for this question because the base is not split into 1.2 and 2.4
Original post by TenOfThem
It is fine to use sine and cosine rules even in right angled triangles

Sadly you do not have the correct answer for this question because the base is not split into 1.2 and 2.4


Ok, its fine, at least i tried...:smile:

What is the answer please explain.
Original post by Dizgurl
Ok, its fine, at least i tried...:smile:

What is the answer please explain.


There is quite a bit more to it as you need to turn the right hand triangle in order to get similar triangles
Original post by TenOfThem
Even assuming that you mean 2.4 instead of 2.6

I do not agree with this premise


Yeah, I meant 2.4 cm. Just wondering then, what would you have done to find out the length of the smaller triangle of the 2? As in, how would you explain that the length is 1.2 cm?
Original post by Chittesh14
Yeah, I meant 2.4 cm. Just wondering then, what would you have done to find out the length of the smaller triangle of the 2? As in, how would you explain that the length is 1.2 cm?


That length is not 1.2
Original post by TenOfThem
That length is not 1.2


In the previous post, i meant "m" not "cm" and what is it then? :/. I want to know where I went wrong.
Original post by Chittesh14
In the previous post, i meant "m" not "cm" and what is it then? :/. I want to know where I went wrong.


It isn't 1.2 at all

Your idea that the 3.6 needed to be in the ratio 1:2 is incorrect

Those two sides are not similar sides
Original post by TenOfThem
It isn't 1.2 at all

Your idea that the 3.6 needed to be in the ratio 1:2 is incorrect

Those two sides are not similar sides


Oh right. Okay, but then what would the answer to it be?
My answer is correct, but the formula is wrong. So then, what would the actual formula be & what is the length of the side 'BC'?
Original post by Chittesh14
Oh right. Okay, but then what would the answer to it be?
My answer is correct, but the formula is wrong. So then, what would the actual formula be & what is the length of the side 'BC'?


Your answer is not correct

You can either use Pythagoras or Harmonic Form to solve the problem

It would be interesting to see where the question came from
Original post by TenOfThem
Your answer is not correct

You can either use Pythagoras or Harmonic Form to solve the problem

It would be interesting to see where the question came from


OH MY GOD! CAN YOU PLEASE EXPLAIN THE ANSWER YOU GOT. Stop saying to chittesh and me we got it wrong. we get it we're wrong!!!
please explain how you did it.

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