ABCD is a square and point P inside the square is such that PCD is an equilateral triangle. Find the angle α
15 DEGREES height of the equilateral triangle is 0.866 times the base. let the base be unity so the height is 0.866 leaving the perpendicular distance from P to AB as 1-.866=0.134 so that alpha is tan inv(0.134/0.5) or 15 degrees.
Alpha is equal to 15 degree Suppose let the side of equilateral triangle be x and this square has also side lengths equal to x . �DPA is isosceles as DP=DA =x. Now In triangle DPA 2(90-alpha) + 30 = 180 Therefore solving we get Alpha equals to 15
PC=BC angle PCB = 30 angle CBP = angle CPB = 1/2(180-30)=75 angle APB = 360 - (75+60+75) = 360 - 210 = 150 angle α = 1/2(180 - 150) = 30/2 = 15
Point P is selected uniformly at random in the interior of square ABCD. What is the probability that angle APD is obtuse (greater than 90 degrees)?
Let ABC be an equilateral triangle. Let P be any point on its incircle. Prove that:
AP2 + BP2 + CP2 = k for some constant k
The square ABCD has a side length of 1, and AEF is an equilateral triangle. What is the area of AEF?
