# Find the lengths of the sides of the outer triangle

1. Jul 12, 2017

### zak100

1. The problem statement, all variables and given/known data
What are the lengths of sides NO and OP in triangle NOP?

See the attached figure

2. Relevant equations
I don’t think that the inner triangle is the 30-60-90 triangle so we cant use the eq of 30-60-90 triangle. However we can use the pythagorous formula to determine the length of hypotnuse of inner triangle. Let X be the other end of hypotenuse of inner triangle:

(40)2+ (24)2 = XP2

So XP = 46.64

3. The attempt at a solution
Sorry I cant figure out how to attempt this question.

Somebody please guide me how to solve it.

Zulfi.

Last edited by a moderator: Jul 12, 2017
2. Jul 12, 2017

### magoo

Isn't there a ratio involved with similar triangles?

Look at the base. One is 40; the other is 40+10.

The ratio of bases is 40/50. You can apply that to other sides.

3. Jul 12, 2017

### Staff: Mentor

The problem doesn't ask for XP. It asks for NO, the vertical side on the left, and OP, the hypotenuse of the larger triangle.
Follow @magoo's recommendation -- these are similar triangles.

4. Jul 13, 2017

### Staff: Mentor

It's much simpler to use the properties of similar triangles -- one can find the length of ON without even writing anything down. Trig isn't needed at all.

5. Jul 13, 2017

### Skins

One of the beautiful things about Mathematics is that there are often several ways of describing a problem or a solution and they imply the same thing yet each way can be uniquely enlightening, . In this case I am applying the properties of similar triangles just as you suggest, namely the tangent is going to remain the same in both similar triangles since the ratio of the opposite and adjacent side of similar triangles will be the same. My solution doesn't require writing anything down or doing any trigonometry to reach the answer. It only serves to illustrate why the property of similar triangles works, because the ratios between the lengths of sides remains the same.

6. Jul 13, 2017

### Ray Vickson

Mod note: The post referred to below has been deleted for the reason given here.
This is coming very close to a complete solution, which is something that PF helpers are forbidden. Telling the poster to use "similar triangles" is one thing, but going through it step-by-step is something we are not supposed to do.

Last edited by a moderator: Jul 13, 2017
7. Jul 14, 2017

### Skins

Okay, fair enough. My bad. I came too close to an actual solution. In retrospect may I rephrase and provide the hint "Consider whether or not the tangent of the angle at P is the same for both triangles and if so what does that imply about the sides of each triangle".

In the future I'll take care to offer helpful hints and avoid explanations in which the limit of explanation approaches an exact solution. :)

8. Jul 17, 2017

### zak100

Hi,
Thanks my friends for your responses. I did not check the earlier close solution which has now been deleted. First i found XP. Note X is a point on the hypotnuse OP intersected by the altitude of smaller triangle and it was 46.64. After that i read the post of magoo. Thanks to him. Then i started reading the theory from the book and i was able to make a following relationship: (Note AP is the base of smaller triangle, AX is the altitude and XP is the hypotenuse of smaller triangle.) So
OP/XP = NP/AP
OP/46.64 = 50/40
OP = 58.3.
Now
(58.3)^2 = (50) ^2 + (NO)^2
(NO)^2 = 898
NO = 29.98

Thanks everybody.

Zulfi.

9. Jul 27, 2017

### I love physics

It can be easily done using trigonometry, and similar triangles. PM me if you still have trouble.