Need Urgent help with Geometry problem!!!!!!!!!!!!!!! (1 Viewer)

[MitDac]

Hi everyone,

I'm new here and it's very nice to meet all of you. I have this problem which seems to be easy at first but turned out to be nasty:burn: Maybe there's an extremely easy way to solve it but it just doesn't come to my mind Here it is:

Triangle ABC is isosceles where AB = AC. angle BAC = 80 degrees. D is a point inside the trianle so that angle BCD = 30 degrees and angle CBD = 10 degree. Find angle BAD!

This is a maths problem for year 7 in Vietnam, so I think it should not be too hard and does not involve complicated formular. Guys, please help! Thanks so much in advance.

 

[jyu]

Hi everyone,

I'm new here and it's very nice to meet all of you. I have this problem which seems to be easy at first but turned out to be nasty:burn: Maybe there's an extremely easy way to solve it but it just doesn't come to my mind Here it is:

Triangle ABC is isosceles where AB = AC. angle BAC = 80 degrees. D is a point inside the trianle so that angle BCD = 30 degrees and angle CBD = 10 degree. Find angle BAD!

This is a maths problem for year 7 in Vietnam, so I think it should not be too hard and does not involve complicated formular. Guys, please help! Thanks so much in advance.

Extend CD until it meets AB at E

It can be shown that ACE and BDE are isosceles triangles.

Let AB = AC = 1 (without loss of generality), and let AE = a and BD = b.

It can be shown that a = 2 sin10deg.

.: BE = ED = 1 - a = 1 - 2sin10

.: b = 2(1 - 2sin10)sin50 = 1.

.: triangleABD is also isosceles.

Since angleABD = 40, .: angleBAD = 70deg.

You need to draw the diagram and mark down all given (and derived) angles to understand.

:wave:

 

[ronaldinho]

this is not a year 7 question...

 

[jyu]

Extend CD until it meets AB at E

It can be shown that ACE and BDE are isosceles triangles.

Let AB = AC = 1 (without loss of generality), and let AE = a and BD = b.

It can be shown that a = 2 sin10deg.

.: BE = ED = 1 - a = 1 - 2sin10

.: b = 2(1 - 2sin10)sin50 = 1.

.: triangleABD is also isosceles.

Since angleABD = 40, .: angleBAD = 70deg.

You need to draw the diagram and mark down all given (and derived) angles to understand.

:wave:

b = 2(1 - 2sin10)sin50 = 1 found with a calculator.

Without calculator:

Use the sine rule to find a = sin20 / sin80

Use the sine rule to find b/sin100 = (1-a)/sin40

b = sin100(1 - sin20 / sin80) /sin40

Since sin100 = sin80, .: b = (sin80 - sin20)/sin40

Use trig formula, b = 2sin30cos50/sin40

Since cos50 = sin40, b = 2sin30 = 1.

:wave:

 

[MitDac]

Thanks for your quick replies, guys. I know this may sound like a joke to you and I wish it is a joke, but it's not. This problem IS for year 7 in Vietnam. My sister in law ask me this question and I can't explain to her using cos or sine rule:burn: I can use your solutions as a hint to find a simple way. Thanks

 

[Archman]

Okey here's a way with no trig:

construct an equilateral triangle with AB as one of the basis, call it ABE. E should lie just outside of triangle ABC, near C.

(in the following bit, whenever i say something like ABC, i mean angle ABC unless specified otherwise)

by construction: AB=AC=BE=AE
now CBE=ABE-ABC=60-50=10=CBD (angle)

EAC=BAC-BAE=80-60=20. Since triangle ACE is isosceles, AEC=ACE=(180-20)/2=80
ECB=ECA-BCA=80-50=30=DCB (angle)

BC is common (side)

so triangle CBD is congruent to triangle CBE (AAS)
so BD=BE (corresponding sides)
=BA

So triangle BAD is isosceles, hence BAD=(180-40)/2=70.

Hope that's not too hard to explain to your sister.

 

[jyu]

Okey here's a way with no trig:

construct an equilateral triangle with AB as one of the basis, call it ABE. E should lie just outside of triangle ABC, near C.

(in the following bit, whenever i say something like ABC, i mean angle ABC unless specified otherwise)

by construction: AB=AC=BE=AE
now CBE=ABE-ABC=60-50=10=CBD (angle)

EAC=BAC-BAE=80-60=20. Since triangle ACE is isosceles, AEC=ACE=(180-20)/2=80
ECB=ECA-BCA=80-50=30=DCB (angle)

BC is common (side)

so triangle CBD is congruent to triangle CBE (AAS)
so BD=BE (corresponding sides)
=BA

So triangle BAD is isosceles, hence BAD=(180-40)/2=70.

Hope that's not too hard to explain to your sister.

An excellent approach based on simple geometry. More suited to clever year 7 students.

One has to think outside the triangle!

:wave:

 

[Affinity]

OMG archman is still here! how are you mate?

 

[Slidey]

Wow. Good to see you back, Ivan.

 

[Archman]

Still browse here sometimes when I'm bored, very rarely though. Also I wasn't going to pass on a nice, cool, decent, interesting (ok I'll stop now) Euclidean geometry problem(ie no trig etc) , don't see too many of those around here.

Oh and another note about that question, its actually quite hard to come up with a nice solution without knowing the answer being 70. So drawing a nice big diagram with ruler and protractor is always good.

 

[MitDac]

Thanks Archman for the neat and simple solution. I've got the answer from her teacher yesterday but only check this thread today. The answer is exactly the same. It's easy to understand but I must admit it's quite hard to think of. Personally, I don't think this problem does any good to a year 7 unless she wants to participate in major Maths competitions.

 

[Slidey]

That is wrong.

It is always important to expose kids to material above their level; if only briefly. They may surprise you with what they can do, and if they don't it may still spur them on to work harder. It's called a challenge; something the education system can't grasp the meaning of these days.