# What is the solution to this tricky triangle problem?

• climbhi
In summary, the conversation discusses a problem found on the web that involves finding angles in a figure. Participants share their solutions and methods, with one using linear algebra and another using the sine and cosine rules. The final answers are determined to be <BCA = 60° and <DBC = 10°, but there is some disagreement over the possibility of parallel lines in the figure.
climbhi
I found a neat problem on the web you can look at it http://physics.harvard.edu/undergrad/prob14.pdf I've always had fun with these find the angle problems but my solutions on this one seemed a little bit weird. Anyone want to let me know what they got? Heres what I found angle BCA = 20 and angle DBC = 50 (the geometry sure doesn't look that way, but I can't find an error in the way I did it and everything seems to be self consistent with my answers...)

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I got the same thing. I'ld forgotten how to get those last two angles because I haven't done anything like that for a few years but when I thorght about all of the infomation that I knew I got the same awnsers.

climbhi, that's a nasty one
I spent some time trying to solve it by the old 180° rule, but didn't succeed. Probably, trig is needed. Being too lazy, I just graphed it. The answer is probably
<BCA = 60°, <DBC = 10°.
Anybody know a nice analytical way to solve it?

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Well, it's easy enough to prove they have to add up to 70. However, the reason you are having trouble is because, for all the triangles in the diagram, they both appear in the same triangles.

So the solution is to create new triangles! My suggestion is to draw a line through C parallel to AD, and see what you can do.

Originally posted by Hurkyl
Well, it's easy enough to prove they have to add up to 70. However, the reason you are having trouble is because, for all the triangles in the diagram, they both appear in the same triangles.

So the solution is to create new triangles! My suggestion is to draw a line through C parallel to AD, and see what you can do.
Yeah drawing new triangles was critical to getting the answers that I got. It's nice to see enslam got the same thing. The way I did it was to draw two new triangles branching off what was already there from which I was able to get a system of four equations with four unknowns, and had my calculator solve from there. The reason I'm uncertain is the way it looks on the page it sure doesn't look like angle DBC is greater then angle BCA.

I don't know so much about linear algebra so maybe the system that I had wasn't fully determined (could very well be the wrong vocab word, I really don't know too much about solving systems with lin. alg.) which lead to there being an answer to the system which was not physically possible?

I'm certain these equations are right, perhaps someone who knows more linear algebra then I do could look at them and tell me if what I suspected up above is right:
x + y = 70
x + b = 50
a + b = 90
y + a = 110

Now I look at it agin the awnser I gave can only be true if DA is parallel to CB and DC is parallel to AB. In this case its not because CDA = 100 and DAB = 100, which means its impossible for DC to be parallel to AB. Thus my awnser was incorrect.

Originally posted by enslam
Now I look at it agin the awnser I gave can only be true if DA is parallel to CB and DC is parallel to AB. In this case its not because CDA = 100 and DAB = 100, which means its impossible for DC to be parallel to AB. Thus my awnser was incorrect.
Hmm well you and I got the same answers but I can't see why those answers are dependant on those criteria. So far I think I've checked every possible quadrilateral and triangle in the thing and the angle measures in them all sum up to 360 and 180 respectively, so the answers appear to be self consistent...

Originally posted by arcnets
climbhi, that's a nasty one
I spent some time trying to solve it by the old 180° rule, but didn't succeed. Probably, trig is needed. Being too lazy, I just graphed it. The answer is probably
<BCA = 60°, <DBC = 10°.
Anybody know a nice analytical way to solve it?

Use the sine rule and the cosine rule to calculate the length between each of the lines. With this info you can find out what the able to find out what the missing angles are.

arcnets was right. The awnser is <BCA = 60°, <DBC = 10°.

What I was getting at with the parallel line is that when a parallel line is bisected with another line the angles are the same. I can't really explain it any better but if u draw two parallel lines and bisect then you'll see what I'm getting at. <ADB and DBC were the same but that can only occur when AB and DC are parallel (stated earlier). Because they weren't parallel then the angles couldn't be the same.

## 1. What is a triangle?

A triangle is a shape that is formed by three straight lines that intersect at three points. It has three sides, three angles, and the sum of its angles is always 180 degrees.

## 2. How many types of triangles are there?

There are three types of triangles: equilateral, isosceles, and scalene. An equilateral triangle has three equal sides and three equal angles. An isosceles triangle has two equal sides and two equal angles. A scalene triangle has no equal sides or angles.

## 3. How do you find the area of a triangle?

The formula for finding the area of a triangle is (base x height) / 2. The base is the length of one side of the triangle, and the height is the perpendicular distance from the base to the opposite vertex. Once you have these measurements, plug them into the formula to find the area.

## 4. What is the Pythagorean theorem?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as a^2 + b^2 = c^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

## 5. How can you determine if three given side lengths form a triangle?

To determine if three given side lengths form a triangle, you can use the triangle inequality theorem. It states that the sum of any two sides of a triangle must be greater than the third side. So, if the sum of the two smaller sides is greater than the longest side, then the three lengths can form a triangle.

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