Triangle ABC is right angled at C. m<ABC=60 and AB=10. Let P be a point randomly chosen inside triangle ABC, such that BP extended meets CA at D. What is the probability that BD>5root(2)?
oops.. misunderstood the problem. BD>5root2. We get the triangle with sides 5root(3) and 5 Now, for BD=5root(2), we get an isosceles triangle with sides 5 each. For this the angle DBC=45deg. So for the small 15 deg part, we get BD>5root(2) so, prob =15/60=.25 (much larger to nahi hai par)
Yes, thats the correct way, So if we fix CD=5 then for the condition to be satisfied P should lie in tri(ABD) out of possible tri(ABC). Nice method :-)
Is the answer to this 0.18?
ReplyDeleteIf it is, i'll post my approach.
No, the answer is much larger than that
ReplyDeleteoops.. misunderstood the problem. BD>5root2.
ReplyDeleteWe get the triangle with sides 5root(3) and 5
Now, for BD=5root(2), we get an isosceles triangle with sides 5 each. For this the angle DBC=45deg.
So for the small 15 deg part, we get BD>5root(2)
so, prob =15/60=.25 (much larger to nahi hai par)
The answer is 1- root(3)/3
ReplyDeleteI think we are comparing sides here. For BD to be 5sqrt2 CD = 5.
ReplyDeleteFor all values of 5<CD<5root3 the condition is satisfied .Hence probability = (root3-1)/root3
yeah right @ Amit, but we will find it easier if we compare areas and not sides
ReplyDeleteYes, thats the correct way, So if we fix CD=5 then for the condition to be satisfied P should lie in tri(ABD) out of possible tri(ABC). Nice method :-)
ReplyDelete