+41 In triangle \(ABC\), we have \(\angle B = 60^o\), \(\angle C = 90^o\), and \(AB = 2\).
Let \(P\) be a point chosen uniformly at random inside \(ABC\). Extend ray \(BP\) to hit side \(AC\) at \(D\). What is the probability that \(BD \lt \sqrt2\)?
Thank you!!!
Triangle ABC is 30-60-90
AB = 2 BC = AB/2 = 1 AC =sqrt [(AB)² - (BC)²] = 1.732050808
CD = sqrt[(BD)² - (BC)²] = 1
Probability that BD < sqrt(2) = (CD) / (AC) = 0.577350269 or 57.7350269 %
I've corrected my error with red "ink"
Nobody's perfect 
+658 Previous comment deleted. Guest has done the duty of showing work.
+658 The restriction \(BD<\sqrt{2}\) creates a 45 - 45 - 90 triangle WITHIN the 30 - 60 - 90 triangle.
We easily calculate the area of the 30 - 60 - 90 triangle.
Small leg = 1
Long leg = \(\sqrt{3}\)
Area = \(\frac{\sqrt{3}}{2}\)
We easily calculate the area of the 45 - 45 - 90 triangle.
Leg = 1
Leg = 1
Area = \(\frac{1}{2}\)
Now the probability of BD being less than root 2 is the area of the 45 - 45 - 90 triangle divided by 30 - 60 - 90 triangle.
Now you do it!
+41 Yes please don't just give an answer to me, If you are, at least give an explanation too!
+128408 A = 30°
D 2
1
C 1 B = 60°
BC = 1
AC = √3
If BD = √2
Then DC = √[ (2) - 1 ] = 1
The area of triangle ABC = (1/2) BC)(AC) = (1/2)(1)(√3) = √3/2
Triangle DCB will have a max area when BD = √2
And its area = (1/2)(CD)(CB) = (1/2)(1)(1) = 1/2
So......the probability that BD < √2 will occur when P falls inside triangle DCB
And this probability is (1/2) / (√3) /2 = 1 / √3
