Problem Solving

sureshbala
thank you!!!! Now clear! Yes, they really should be 45!!! Now I don not understand how I missed it - actuallym this rule with arcs is very easy
Once again, thank you! I love this site for the opportunity to solve all difficulties! ))

4meonly wrote:sureshbala
thank you!!!! Now clear! Yes, they really should be 45!!! Now I don not understand how I missed it - actuallym this rule with arcs is very easy
Once again, thank you! I love this site for the opportunity to solve all difficulties! ))

Nice to hear that you were able to recollect the basics...anyway...we have a lot to come in this thread....stay tuned.

Regards

I will!!!

Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

sureshbala wrote:

In the above figure, ABC is a right angled triangle. Taking the hypotenuse AC a square is constructed. If O is the center of the square find angle ABO.

I didn't have a proper proof for this inside 2 minutes, but I think I could have guessed it correctly.

If the question was worded as above, then you could reasonably make ABC an isosceles right-angles triangle, without loss of generality. AC is then root-2 times as long as the other sides, and OB clearly bisects the angle ABC.

Similarly, imagine a right-angled triangle where BC is vanishingly small, so AC is only a tiny bit longer than AB. As AB gets smaller, then OB gets closer to OC, which is also 45 degrees.

Not beautiful maths, but enough for a confident guess.

sureshbala wrote:Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

Folks, I hope this is keeping you busy....

sureshbala wrote:

sureshbala wrote:Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

Folks, I hope this is keeping you busy....

I think I have a way to solve this, but it's not something I could reproduce in a test:

Drop a line from P to a new point, Q, on AB, such that LPQB = 90°. Triangle PQB is a right-angled isoscelese triangle, so length PQ = (length PB / root 2).

Now consider triangle PQA, with angles 15-75-90, and one side, PQ, with length given above:
sin 15° = PQ / PA

length PA = (1/(root 2)) / sin 15°

Now draw another new line, this one from C to AP, which it meets at R, such that LCRP = 90°. Triangle CRP is a 30-60-90 triangle, and CP is two units long, in the units we are using. Length CR is therefore (root 3), and length RP = 1.

Finally, triangle CRA is right-angled, and has two sides of length (root 3) and ((1/(root 2)) / sin 15°) - 1). Bizarrely, this second term also turns out to be (root 3), (and yes, I needed a calculator to know that). So it's also an isosceles right-triangle, and angle ACR is therefore 45°. This must be added to angle RCD, previously found to be 30°, to give

ACB = 75°

But like I said, I couldn't do this without a calculator and ten minutes. What's the trick?

sureshbala wrote:angles made by two equal chords on the same segment of the circle are always equal.

Can you explain what this means? Over here:

https://www.craville.110mb.com/maths/circlegeometry.pdf

I read this:

"Angles in the same segment of a circle are equal."
- Sorry, I can't reproduce diagrams, I don't have any kind of drawing tool

I'm not sure what is meant by "segment" in this case? Can you explain it to me, or give me a link to a fuller explanation?

thanks!
James

Never mind, just found it here :

https://www.mathsteacher.com.au/year10/c ... egment.htm

sureshbala wrote:Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

Folks, waiting for some more replies....anyway I will post the solution soon...keep trying

I wonder if the LABC is 90?

LACB is 90. Since LAPC=60, LAPB=120. Since LAPB=120, LPAB=15.
Now you can figure out the angles of triangel ACP.

gxliu wrote:LACB is 90. Since LAPC=60, LAPB=120. Since LAPB=120, LPAB=15.
Now you can figure out the angles of triangel ACP.

Dear gxliu....you were too quick in your conclusion....please check your calculation again.

You're right. LACB is 45. I forgot to take into account the sides. tricky...

gxliu wrote:You're right. LACB is 45. I forgot to take into account the sides. tricky...

hmmm....you have to check again....