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Math Expert V
Joined: 02 Sep 2009
Posts: 62379
Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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sheolokesh wrote:
Hi Bunuel,

please explain me the correlation between 30-60-90(1:sq(3):2), 45-45-90(1:1:sq(2)) with pythagorean triplets(3-4-5, 7-24-25). I am confused because if we have a 90deg triangle, and 2 sides 3,4 we can put other side as 5? If so then it becomes 30-60-90 triangle right? then why we are not able to correlate 1:sq(3):2 with 3:4:5?
because we need to commonly multiply the ratio, so if 1*3 then sq(3) must also multiply by 3 which is not equal to 4..

Because I had tried an another approach which gives a wrong ans... Need clarification.. If ABC is a right triangle with 3,4 then other side must be 5. So Angle BAC will be 30 deg. which makes Ang CAD as 60 and CDA as 30.. So we have 30-60-90 triangle on ACD. If thats the case, then CD must be 4*sqrt(3) right? what's wrong in my approach?

sheolokesh wrote:
if we have a 90deg triangle, and 2 sides 3,4 we can put other side as 5

Yes. If 3 and 4 are legs of a right triangle, then hypotenuse is 5.

sheolokesh wrote:
If so then it becomes 30-60-90 triangle right?

No. The angles in 3-4-5 right triangle are NOT 30°-60°-90°. They are ~53° - ~37 ° - 90°.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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So all triplets has 37-53-90 degrees right? Its better not to confuse triplets with 30-60-90 or 45-45-90 then..
Math Expert V
Joined: 02 Sep 2009
Posts: 62379
Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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sheolokesh wrote:
So all triplets has 37-53-90 degrees right? Its better not to confuse triplets with 30-60-90 or 45-45-90 then..

No. In 3-4-5, 6-8-10, 9-12-15, ... right triangles angles will be ~53° - ~37 ° - 90°. But for example, in 5-12-13 right triangle angles are 23°-67°-90°.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Hey Could any one please tell me if my approach is correct! I used trigonometry!

Approach below:
Consider triangle ABC
cosθ = 3/5 (base/hyp)........(1)
Consider triangle ABD
cosθ = 5/(3+CD).........(2)

Equate (1) and (2)

3/5=5/(3+CD)
3CD+9= 25 ====> CD=16/3
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Quote:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3

As depicted in the figure, AB^2 = 3^2 + 4^2 = 5^2
i.e. AB = 5

NOW TRIANGLE ABC AND TRIANGLE ACD ARE SIMILAR TRIANGLE

So the ratio of their corresponding sides will be equal

AC/BC = CD/AC

4/3 = CD/4

i.e. CD = 16/3

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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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How the angle BAC is equal to angle ADC
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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sudhanshutiwari wrote:
How the angle BAC is equal to angle ADC

sudhanshutiwari
Take bigger triangle ABD and mark Angles as 90-x - (90-x)

Now look at triangle ACD, one of the angle you can already see marked is (90-x) one angle is 90 so third angle becomes x (if you subtract other two angles from 180)

Similarly, look at triangle ABC, one of the angle you can already see marked is x one angle is 90 so third angle becomes 90-x (if you subtract other two angles from 180)

This is how, we figure out that angle BAC is equal to angle ADC

I hope this helps
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Both smaller triangles are similar to the large triangle. So they are similar to each other too.

Angle BAD = BCA (90 degrees)
and angle B is common in both
So by AA, triangles BAD and BCA are similar

angle BAD = ACD (90 degrees)
and angle D is common in both
So by AA, triangles BAD and ACD are similar

So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.

Responding to a pm:
Quote:
Can you explain why it's not BCA is similar to DCA? Since according to the image, A is the same for both triangles and C is the same, so shouldn't it be BCA and DCA?

Note that angles BCA and DCA are both 90 degrees so that's fine but angles BAC and DAC are not equal. These are two distinct angles and can take any values. Using the logic I have discussed in my solution, we find that triangle BCA is similar to triangle ACD.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Both smaller triangles are similar to the large triangle. So they are similar to each other too.

Angle BAD = BCA (90 degrees)
and angle B is common in both
So by AA, triangles BAD and BCA are similar

angle BAD = ACD (90 degrees)
and angle D is common in both
So by AA, triangles BAD and ACD are similar

So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.

Responding to a pm:
Quote:
Can you explain why it's not BCA is similar to DCA? Since according to the image, A is the same for both triangles and C is the same, so shouldn't it be BCA and DCA?

Note that angles BCA and DCA are both 90 degrees so that's fine but angles BAC and DAC are not equal. These are two distinct angles and can take any values. Using the logic I have discussed in my solution, we find that triangle BCA is similar to triangle ACD.

Hi VeritasKarishma. I understood that BAC and DAC are not equal cus we don't know whether the bisector AC is bisecting angle A equally; but how do we establish angle B to be similar to angle A?? did we derive it from the above ratios?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10229
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Kritisood wrote:
Both smaller triangles are similar to the large triangle. So they are similar to each other too.

Angle BAD = BCA (90 degrees)
and angle B is common in both
So by AA, triangles BAD and BCA are similar

angle BAD = ACD (90 degrees)
and angle D is common in both
So by AA, triangles BAD and ACD are similar

So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.

Responding to a pm:
Quote:
Can you explain why it's not BCA is similar to DCA? Since according to the image, A is the same for both triangles and C is the same, so shouldn't it be BCA and DCA?

Note that angles BCA and DCA are both 90 degrees so that's fine but angles BAC and DAC are not equal. These are two distinct angles and can take any values. Using the logic I have discussed in my solution, we find that triangle BCA is similar to triangle ACD.

Hi VeritasKarishma. I understood that BAC and DAC are not equal cus we don't know whether the bisector AC is bisecting angle A equally; but how do we establish angle B to be similar to angle A?? did we derive it from the above ratios?

This is a standard figure that begs you to think about similar triangle. Some such figures are given in this post on Veritas blog: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/

Here, triangle CBA is similar to CDA and both are similar to triangle ABD.
If you want to see how, note this:

Between triangles CBA and ABD,
Angle C is 90 and angle BAD are 90 each. Angle B is common to both. So by AA, both triangles are similar.

Angles ACD and BAD are 90 each. Angle D is common to both. So by AA, both triangles are similar.

Angle C is 90.
Now say angle BAC is x, then angle CAD is (90-x). Then in triangle CAD, angle D must be x.
Hence angle BAC = angle D
So by AA, both triangles are similar.
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Karishma
Veritas Prep GMAT Instructor

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Joined: 21 Feb 2017
Posts: 208
Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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Kritisood wrote:

Responding to a pm:
Quote:
Can you explain why it's not BCA is similar to DCA? Since according to the image, A is the same for both triangles and C is the same, so shouldn't it be BCA and DCA?

Note that angles BCA and DCA are both 90 degrees so that's fine but angles BAC and DAC are not equal. These are two distinct angles and can take any values. Using the logic I have discussed in my solution, we find that triangle BCA is similar to triangle ACD.

Hi VeritasKarishma. I understood that BAC and DAC are not equal cus we don't know whether the bisector AC is bisecting angle A equally; but how do we establish angle B to be similar to angle A?? did we derive it from the above ratios?

This is a standard figure that begs you to think about similar triangle. Some such figures are given in this post on Veritas blog: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/

Here, triangle CBA is similar to CDA and both are similar to triangle ABD.
If you want to see how, note this:

Between triangles CBA and ABD,
Angle C is 90 and angle BAD are 90 each. Angle B is common to both. So by AA, both triangles are similar.

Angles ACD and BAD are 90 each. Angle D is common to both. So by AA, both triangles are similar.

Angle C is 90.
Now say angle BAC is x, then angle CAD is (90-x). Then in triangle CAD, angle D must be x.
Hence angle BAC = angle D
So by AA, both triangles are similar.[/quote]

Oh... understood now.   Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the   [#permalink] 24 Mar 2020, 03:10

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