1. ## Strange Question.

I have tried everything I can think of on this and I must be missing something. I had a student come into the tutoring center with this. The answer is supposed to be 57.1 degrees. It seems like it should be simple, but I can't for the life of me get 57.1 degrees. So either the book is wrong or, like I said, I am missing something. Which is more likely.

2. Originally Posted by Jen
I have tried everything I can think of on this and I must be missing something. I had a student come into the tutoring center with this. The answer is supposed to be 57.1 degrees. It seems like it should be simple, but I can't for the life of me get 57.1 degrees. So either the book is wrong or, like I said, I am missing something. Which is more likely.

I've modified your sketch a little bit (see attachment).

The straight part of the road is tangent to the circle. That means the radius is perpendicular to this part of the road.

Then you are dealing with 2 right triangles. Calculate the angles $\alpha$ and $\beta$.

I got $\alpha + \beta \approx 50.1^\circ$

3. I never thought of that set up. I am starting to think that the book is wrong. The thing is, this question is from a math 53 class, which is a really quick, down and dirty, trig class for building inspection majors. This student only knows basic Sin, Cos, Tan, functions and some area formulas. I don't think that they have ever talked about a line being tangent. Bar that however, I still think the book might have a typo because I have tried everything with every assumption I could think of.

Thank you so much for your help. That was a much better set up than any I had tried

If you come up with anything else please let me know.

4. Originally Posted by Jen
I never thought of that set up. I am starting to think that the book is wrong. The thing is, this question is from a math 53 class, which is a really quick, down and dirty, trig class for building inspection majors. This student only knows basic Sin, Cos, Tan, functions and some area formulas. I don't think that they have ever talked about a line being tangent. Bar that however, I still think the book might have a typo because I have tried everything with every assumption I could think of.

Thank you so much for your help. That was a much better set up than any I had tried

If you come up with anything else please let me know.
It is very early in the morning here and so it could happen that I missed something:

The two right triangles must be congruent: Both have the same hypotenuse and both have a leg of 180. Therefore $\alpha = \beta$.

Therefore the angle in question is $y = 2 \cdot \beta = 2\cdot 26.25^\circ = 52.5^\circ$

It seems to me that I'll reach the correct answer in the next attempt.

5. Originally Posted by earboth
It is very early in the morning here and so it could happen that I missed something:

The two right triangles must be congruent: Both have the same hypotenuse and both have a leg of 180. Therefore $\alpha = \beta$.

Therefore the angle in question is $y = 2 \cdot \beta = 2\cdot 26.25^\circ = 52.5^\circ$

It seems to me that I'll reach the correct answer in the next attempt.

Haha, that is about what I have. If you include the width of the road you can make it as large as 53.7, which can be as close to 57.1 as you wish for very large values of 53.7 right..

6. Originally Posted by Jen
Haha, that is about what I have. If you include the width of the road you can make it as large as 53.7, which can be as close to 57.1 as you wish for very large values of 53.7 right..
I played a little bit with your sketch:

If you cut off 40' from the distance of 365' (which is the wrong method because the width of the road is measured perpendicular from side to side) then you'll get an angle of 57.96°.

Actually you must use the angle $\beta$ to calculate the distance to be cut off the 365' (=44.6') but I don't have the time to do the calculations in detail: I must fetch some rolls for breakfast now.

7. Have a lovely breakfast, and thanks again!

8. Originally Posted by earboth
It is very early in the morning here and so it could happen that I missed something:

The two right triangles must be congruent: Both have the same hypotenuse and both have a leg of 180. Therefore $\alpha = \beta$.

Therefore the angle in question is $y = 2 \cdot \beta = 2\cdot 26.25^\circ = 52.5^\circ$

It seems to me that I'll reach the correct answer in the next attempt.
You were correct!!

I am sure you weren't worried about it but I thought I would let you know.