Question
ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude from A to BC. and P and φ denote the perimeter and area of the triangle respectively, then lim n−» 0 φ/p³ is equal to??
first draw a cirle and triange inside it , consider angle B and angle C asα so angle A = 180-2α, let centre of the circle is O and D is the point in the line BC where altitude meets in the line BC, given that altitude = h and radius = r
so AB = AC = h cosecα and BC = 2 h cotα so
p = 2 h (cosecα + cotα)
and φ = 1/2 2 h cotα h = h2 cotαa
and in triangle OBD angle OBD = 2α - 90 so cos(2α - 90) = h cotα /r implies that h = 2 r sin2α
now i think limit will be based on h not n
lim h−» 0 φ/p³ = 1/128r
for getting solution put p and φ first then put h in terms of r, you will get an equation based on α and r,
convert limh->0 to limα->0 because when h will be 0, α will also be 0