Trigonometry Practice

[AlanHaisley]

O.K. here's a difficult one - its not really a trick question like my earlier triangle and you can work out the answer the hard way (all the formulae are given).
There is a simple method if you apply your mind.
(I did it the hard way before I figured it out).

It would appear from first glance that there is insufficient information to answer the question - but there is.

Ken

Ken,

Nice problem! When I started on this, it seemed to me that - as you say - I needed at least one more piece of information.

Finally, I decided to see what the answer would be if I just tried a couple of sizes of holes; I'm afraid that I was still a bit muzzy minded at the time.

For one hole I picked, of course, the largest possible sphere. Obviously this gave me a drill diameter of zero. It also gave me a sphere diameter of 100mm.

For the other hole I picked one with a diameter the same as the depth of hole. Interestingly, I got the same answer for the volume. Surprise, surprise!

Finally I realized that I could indeed express D, d, and h in terms of L.

I decided not to post my work here in case others are playing around with this; still, my hints should lead others to the solution.

Even having worked this out, the result is counter-intuitive.

 

[Ken I]

Counter intuitive indeed, the answer is a constant - hardly credible really - glad you liked it.

I posted it before some time back but no one "bit"

Ken

 

[JohnT]

Given a circle of radius R with a chord with a length of 26in. At the midpoint of the chord there is a perpendicular bisector that intersects the circle and has a length of 3.75in. Find R.
This is a real world problem that came up once or twice in my career. It can be solved by 3 simultaneous equations in 3 unknowns or reduced row echelon form of a matrix , but there a couple of ways to solve it directly far faster.

 

[Maryak]

R = (c^2+4h^2)/8h

R = (676+56.25)/30

R = 24.4083333

 

[Ken I]

I got the same answer as Bob from Pythagoras - ?

Ken