The answer is \(78^\circ\), not \(80^\circ\). In a problem like this, one approach is to construct the point another way, and then show that it has the same defining properties as the point given in the problem.

Let \(E\) be the point inside triangle \(ABC\) such that \(AE = AC\) and \(\angle CAE = 24^\circ\). Let \(F\) be the point inside triangle \(ACE\) such that triangle \(CEF\) is equilateral.

What can you say about triangles \(AFC\) and \(BEC\)? What does that say about points \(D\) and \(E\)?
–
Jon Haussmann
·
3 years, 3 months ago

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@Jon Haussmann
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Brilliant! and then the angle of the bases of the isosceles triangles are both 78! Amazing!
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Finn Hulse
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3 years, 3 months ago

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There is one more (more geometric, no trigonometry) solution. Let us costruct the rombus ACBE. The angle AEB will also be 96 degrees. As the sum of the angles DAB and ABD is 48 degrees, then the point D will occur on the circumference with the center E. And AE=ED. The angle DAE equals 60 degrees, then EA=ED=AD. And AC=AD and we come to the same result - ACD equals 78 degrees.
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Сергей Кротов
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3 years, 3 months ago

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The problem is not at all tough.. If you are versed with sine law then it isnt even a three step problem
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Aarsh Verdhan
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2 months, 2 weeks ago

Dear Finn Hulse. The answer is 78 degrees. To get this you should prove that AD=AC. In my solution I considered two triangles ACK (K lies in the center of AB) and ADB. Then AK=AB/2=ACsin48 and AD/sin30=AB/sin48 (a sine-theorem). You then directly get AC=AD.
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Сергей Кротов
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3 years, 3 months ago

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dear Mr. Finn Hulse
actually there are missing data in the problem. i tried to solve it many times but always the problem redduces to to the following simultaneous equations
x+y = 228
y+z = 168
x- z = 60
At the end of the day , if you can get a solution to these equations and find the value of z ,
your required angle is ( 96 - z ) . . . . .now I think the problem is how to find z ?
thanks
yours : aziz alasha
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Aziz Alasha
·
3 years, 3 months ago

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@Aziz Alasha
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Same here! Apparently you have to start drawing lines to solve it.
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Finn Hulse
·
3 years, 3 months ago

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http://math.stackexchange.com/questions/245608/find-an-angle-in-a-given-triangle
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Shamik Banerjee
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3 years, 3 months ago

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If you can make it on the wall, you can also make it on paper with protractor and scale and can landed on right answer. That's what i did and the correct answer is 80 degrees.
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Raghav Garg
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3 years, 3 months ago

@Raghav Garg
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Conor is correct. I don't like to use protractors to solve math. But thank you so much for taking the time to solve this! It's very much appreciated. Now all I need to know is why.
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Finn Hulse
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3 years, 3 months ago

## Comments

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TopNewestThe answer is \(78^\circ\), not \(80^\circ\). In a problem like this, one approach is to construct the point another way, and then show that it has the same defining properties as the point given in the problem.

Let \(E\) be the point inside triangle \(ABC\) such that \(AE = AC\) and \(\angle CAE = 24^\circ\). Let \(F\) be the point inside triangle \(ACE\) such that triangle \(CEF\) is equilateral.

http://i.imgur.com/Fb0iir4.png

What can you say about triangles \(AFC\) and \(BEC\)? What does that say about points \(D\) and \(E\)? – Jon Haussmann · 3 years, 3 months ago

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– Finn Hulse · 3 years, 3 months ago

Brilliant! and then the angle of the bases of the isosceles triangles are both 78! Amazing!Log in to reply

There is one more (more geometric, no trigonometry) solution. Let us costruct the rombus ACBE. The angle AEB will also be 96 degrees. As the sum of the angles DAB and ABD is 48 degrees, then the point D will occur on the circumference with the center E. And AE=ED. The angle DAE equals 60 degrees, then EA=ED=AD. And AC=AD and we come to the same result - ACD equals 78 degrees. – Сергей Кротов · 3 years, 3 months ago

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The problem is not at all tough.. If you are versed with sine law then it isnt even a three step problem – Aarsh Verdhan · 2 months, 2 weeks ago

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I proved it. – Ashu Pateriya · 1 year, 10 months ago

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Dear Finn Hulse. The answer is 78 degrees. To get this you should prove that AD=AC. In my solution I considered two triangles ACK (K lies in the center of AB) and ADB. Then AK=AB/2=ACsin48 and AD/sin30=AB/sin48 (a sine-theorem). You then directly get AC=AD. – Сергей Кротов · 3 years, 3 months ago

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dear Mr. Finn Hulse actually there are missing data in the problem. i tried to solve it many times but always the problem redduces to to the following simultaneous equations x+y = 228

y+z = 168 x- z = 60 At the end of the day , if you can get a solution to these equations and find the value of z , your required angle is ( 96 - z ) . . . . .now I think the problem is how to find z ? thanks yours : aziz alasha – Aziz Alasha · 3 years, 3 months ago

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– Finn Hulse · 3 years, 3 months ago

Same here! Apparently you have to start drawing lines to solve it.Log in to reply

http://math.stackexchange.com/questions/245608/find-an-angle-in-a-given-triangle – Shamik Banerjee · 3 years, 3 months ago

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If you can make it on the wall, you can also make it on paper with protractor and scale and can landed on right answer. That's what i did and the correct answer is 80 degrees. – Raghav Garg · 3 years, 3 months ago

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– Conor Sokolowsky · 3 years, 3 months ago

He doesn't have a protractor. But can you prove it?Log in to reply

– Finn Hulse · 3 years, 3 months ago

Conor is correct. I don't like to use protractors to solve math. But thank you so much for taking the time to solve this! It's very much appreciated. Now all I need to know is why.Log in to reply