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?
80% of success is showing up -- Woody Allen
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