# Trig ratio question (picture included)

1. Apr 15, 2012

### zeion

1. The problem statement, all variables and given/known data

Hi,

How do I solve for y in this triangle? (picture in attachment)

2. Relevant equations

So I use the special angel ratio and get

sqrt(3)x = y + 20

and then from the smaller triangle I get

tan(theta) = 20 / x

What do I do next?

Thanks.

3. The attempt at a solution

#### Attached Files:

• ###### drawing.pdf
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2. Apr 15, 2012

### Mentallic

There doesn't seem to be enough information to solve explicitly for x and y. The best you can do is have a relationship between x and y, which you've already found.

3. Apr 15, 2012

### Robert1986

Let $h$ be the length of the hypotenuse. Then $(20+y)^2 + x^2 = h^2$. But, since the big triangle is a 60-30-90 triangle, you can write $x$ and $y$ in terms of $h$ in the equation $(20+y)^2 + x^2 = h^2$. (As a hint: $x = h/2$.

Now, you have a quadratic in $h$ which you can solve for.

4. Apr 15, 2012

### zeion

So I'm supposed to write both x and y in terms of h?

5. Apr 15, 2012

### Vorde

This is correct, assuming the 60 degrees in just for the inner angle, not the larger angle.

6. Apr 15, 2012

### Robert1986

Yes, but this is assuming that the 60 degrees is for the big angle, not the small angle. If the 60 is for the small angle, then you are in trouble because I can't see a way to figure it out.

7. Apr 15, 2012

### Mentallic

60o is most likely the larger angle because of its positioning and the fact that the angle illustration doesn't stop at the hypotenuse of the smaller triangle.

zeion, your first answer of sqrt(3)x = y + 20 was correct, and it's the best you can do right now, unless some other info was given, such as the length of the larger hypotenuse, or one of the smaller angles.

8. Apr 16, 2012

### Robert1986

I agree the 60 degrees is the angle of the bigger angle; I disagree that it can't be solved.

Let $h$ be the length of the hypotenuse. Then $x=h/2$ and $20+y=\sqrt{3}h/2$ Additionally, $(20+y)^2 + x^2 = h^2$. Now, we can write $h^2/4 + 3h^2/4 = h^2$. So, we know the hypotoneuse which means we can find $y$.

EDIT:

Here is my concern: if the 60 degrees is the angle of the bigger angle (as I assumed) then what even is the point of the inner triangle? Zeion, is there another problem that uses the inner triangle?

9. Apr 16, 2012

### Mentallic

But what you've done is essentially shown that the pythagorean theorem is true using trigonometric values. You haven't specifically solved for anything, which is what we were trying to do (and can't do).

So in other words, yes, since we have that $20+y=\frac{\sqrt{3}h}{2}$ and $x=\frac{h}{2}$ then we can show that $(20+y)^2 + x^2 = h^2$ is true, but we haven't found a value for h.
No we don't :tongue:

Exactly. I'm still waiting for that 1 piece of extra information that we need to solve the problem.

10. Apr 16, 2012

### Robert1986

Which one is the embarrassed face? I need that one. I made a sign error in my calculation (not the ones I posted, by the ones I did on my own paper). In fact, the sign error I made implies that $h^2 = -3/4$ or something. As I see now, I have proven the astonishing fact that $h^2 = h^2$, and to think I thought that geometry wasn't my forte!

11. Apr 16, 2012

### Mentallic

I tend to use but could also work I guess. Making a mistake in maths is the same as being nervous to ask a girl out, right?

Then h is imaginary!? :tongue:

No worries, it was a fair mistake. Also, since you stopped at that point where you said
"So, we know the hypotoneuse which means we can find y."
I'm guessing you were probably unsure what was going on with the equation for your hypotenuse h?

Oh and I think you were meant to say that you thought geometry was your forte haha.

12. Apr 16, 2012

### zeion

I may have copied the question wrong; will update once I find out :/

13. Apr 22, 2012

### zeion

The questions says that the middle line (the one inside the big triangle that cuts it into two smaller triangle) bisects the altitude / height of the triangle (connects to the middle of y +20).
Would that mean that the smaller triangle has a 30 deg angle?

14. Apr 22, 2012

### Mentallic

Ahh, our final piece of information.

If it bisects the line, then no it doesn't mean the smaller angle is half the size of the larger. If it were, computing half angles in trigonometry would be so much easier :tongue:

Think about an extreme example: imagine the line y+20 is so large that the larger angle is nearly 90o, but a line going up half way would still have a near 90o angle, as opposed to a 45o angle.

To finish off the problem, denote the smaller angle $\theta$. Now what is $\tan(\theta)$ equal to?

15. Apr 22, 2012

### zeion

Okay I'm confused. Let's label the corners of the triangle as A, B for the bottom (left to right) and C for the top, and the middle line as AD. So if AD bisects CB, doesn't that mean that y is also 20?

16. Apr 22, 2012

### Mentallic

I should have looked back at the diagram. I thought the side was labelled y+20 (which would be weird now that I think about it) as opposed to each section being labelled y and 20 respectively.

Yes, y is 20.

Also, I'm curious. If the information of AD bisecting BC was given to you, what is the point of all the other labels? To put you off? Or is this a part of a bigger problem?