Trig ratio question (picture included)

In summary, the student is trying to solve for y in a triangle. They found that (20+y)^2 + x^2 = h^2 and that x = h/2. They are trying to find y from this information. However, they are having difficulty because they don't have explicit information about x and y. The student has found that h^2 = h^2, but they are still missing a value for h.
  • #1
zeion
466
1

Homework Statement



Hi,

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


Homework 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.

The Attempt at a Solution

 

Attachments

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  • #2
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
Let [itex]h[/itex] be the length of the hypotenuse. Then [itex](20+y)^2 + x^2 = h^2[/itex]. But, since the big triangle is a 60-30-90 triangle, you can write [itex]x[/itex] and [itex]y[/itex] in terms of [itex]h[/itex] in the equation [itex](20+y)^2 + x^2 = h^2[/itex]. (As a hint: [itex]x = h/2[/itex]. Now, you have a quadratic in [itex]h[/itex] which you can solve for.
 
  • #4
Robert1986 said:
Let [itex]h[/itex] be the length of the hypotenuse. Then [itex](20+y)^2 + x^2 = h^2[/itex]. But, since the big triangle is a 60-30-90 triangle, you can write [itex]x[/itex] and [itex]y[/itex] in terms of [itex]h[/itex] in the equation [itex](20+y)^2 + x^2 = h^2[/itex]. (As a hint: [itex]x = h/2[/itex].


Now, you have a quadratic in [itex]h[/itex] which you can solve for.

So I'm supposed to write both x and y in terms of h?
 
  • #5
Mentallic said:
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.

This is correct, assuming the 60 degrees in just for the inner angle, not the larger angle.
 
  • #6
zeion said:
So I'm supposed to write both x and y in terms of h?

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
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
Mentallic said:
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.

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

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

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
Robert1986 said:
I agree the 60 degrees is the angle of the bigger angle; I disagree that it can't be solved.

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

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 [itex]20+y=\frac{\sqrt{3}h}{2}[/itex] and [itex]x=\frac{h}{2}[/itex] then we can show that [itex](20+y)^2 + x^2 = h^2[/itex] is true, but we haven't found a value for h.
Robert1986 said:
So, we know the hypotoneuse which means we can find [itex]y[/itex].
No we don't :tongue:

Robert1986 said:
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?
Exactly. I'm still waiting for that 1 piece of extra information that we need to solve the problem.
 
  • #10
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 [itex]h^2 = -3/4[/itex] or something. As I see now, I have proven the astonishing fact that [itex]h^2 = h^2[/itex], and to think I thought that geometry wasn't my forte!
 
  • #11
Robert1986 said:
Which one is the embarrassed face? I need that one.
I tend to use :redface: but :blushing: could also work I guess. Making a mistake in maths is the same as being nervous to ask a girl out, right?

Robert1986 said:
In fact, the sign error I made implies that [itex]h^2 = -3/4[/itex] or something.
Then h is imaginary!? :tongue:

Robert1986 said:
As I see now, I have proven the astonishing fact that [itex]h^2 = h^2[/itex], and to think I thought that geometry wasn't my forte!
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
I may have copied the question wrong; will update once I find out :/
 
  • #13
Hi! Thread necro!
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
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 [itex]\theta[/itex]. Now what is [itex]\tan(\theta)[/itex] equal to?
 
  • #15
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
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?
 

1. What are trigonometric ratios?

Trigonometric ratios are mathematical functions that relate the angles and sides of a right triangle. They are used to solve for missing side lengths or angles in a triangle.

2. What are the three main trigonometric ratios?

The three main trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). These ratios are calculated by dividing the length of a specific side of a right triangle by another side.

3. How do I use trigonometric ratios to solve a problem?

To solve a problem using trigonometric ratios, you first need to identify the given information, such as the length of known sides or angles. Then, choose the appropriate ratio and set up an equation using the given information. Finally, solve for the unknown value using algebra.

4. What is the unit circle and how does it relate to trigonometric ratios?

The unit circle is a circle with a radius of 1 unit. It is used to visualize and understand the trigonometric ratios. The coordinates of points on the unit circle correspond to the values of sine, cosine, and tangent for specific angles.

5. Can I use trigonometric ratios for any type of triangle?

No, trigonometric ratios can only be used for right triangles. However, they can be applied to any right triangle, regardless of its size or shape.

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