I am attempting to find the distance between the two arrows (the short side of the small right triangle) in this figure. I'm given that the area of circle = 2 x (area of the square) = 7590. From that information, I've calculated that the length of a side of the square is approximately 61.6 and the length of the radius of the circle is approximately 49.2. I know that the tangent lines are perpendicular to radii of the circle, thus forming additional right triangles. I can't figure out what other relationships I can use to find the solution.
circle, tangent lines, square, right triangles,......
- Thread starter calcgal
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The little side of the triangle can be expressed in two ways. Assuming the tangent at the square is halfway up the square. That's how it looks.
I reckon that is a safe assumption.
By the little triangle, let a=the side we need.
This can be expressed as \(\displaystyle a=61.6cot(\theta)\)
Another way to express it is \(\displaystyle a=49.15tan(\frac{\theta}{2})-30.8\)
Set these equal and we get an angle of \(\displaystyle 80.228269333\) degrees.
Therefore, \(\displaystyle a=61.6cot(80.228269333)=10.6\)
Another way to look at it is noting that the angle at the top of the little triangle is the same as the central angle in the circle between the tangent lines. This is called the Delta angle. That is what they called it when I surveyed. One could use this angle to derive some needed info.
That's what I was getting at a while ago and made a mistake.
Here is a graph to show what I mean. The distance between A and D is called the External. It is derived by using an old-timey, obsolete, trig function called the exsecant. Soroban and SK probably remember it
I reckon that is a safe assumption.
By the little triangle, let a=the side we need.
This can be expressed as \(\displaystyle a=61.6cot(\theta)\)
Another way to express it is \(\displaystyle a=49.15tan(\frac{\theta}{2})-30.8\)
Set these equal and we get an angle of \(\displaystyle 80.228269333\) degrees.
Therefore, \(\displaystyle a=61.6cot(80.228269333)=10.6\)
Another way to look at it is noting that the angle at the top of the little triangle is the same as the central angle in the circle between the tangent lines. This is called the Delta angle. That is what they called it when I surveyed. One could use this angle to derive some needed info.
That's what I was getting at a while ago and made a mistake.
Here is a graph to show what I mean. The distance between A and D is called the External. It is derived by using an old-timey, obsolete, trig function called the exsecant. Soroban and SK probably remember it
