Find the lengths of the sides of the outer triangle

[zak100]

Homework Statement

[/B]What are the lengths of sides NO and OP in triangle NOP?

See the attached figure

Homework Equations

[/B]I don’t think that the inner triangle is the 30-60-90 triangle so we can't 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

The Attempt at a Solution

Sorry I can't figure out how to attempt this question.

Somebody please guide me how to solve it.

Zulfi.

 

[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.

 

[Mark44]

Homework Statement

[/B]What are the lengths of sides NO and OP in triangle NOP?

See the attached figure

Homework Equations

[/B]I don’t think that the inner triangle is the 30-60-90 triangle so we can't 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

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.

The Attempt at a Solution

Sorry I can't figure out how to attempt this question.

Somebody please guide me how to solve it.

Zulfi.View attachment 207030

Follow @magoo's recommendation -- these are similar triangles.

 

[Mark44]

First considering the "inner triangle" The tangent of the angle at P is 24/40, i.e. tan(angle at P) = 24/40. So we know the tangent of that angle and that tangent has to be the same for the "outer triangle" as well. So tan(angle at P) = 24/40 = x/(40+10) or 24/40 = x/50. Solve for x and you'll have the length OP. So now you have 2 sides of a triangle and now it's trivial to find the hypotenuse OP. (hint - Pythagorian theorem)

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.

 

[Skins]

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.

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.

 

[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.

 

[Skins]

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.

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. :)

 

[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

Answers are correct.
Thanks everybody.

Zulfi.

 

[I love physics]

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