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Check below? (geometry involved)

Question

Check below? (geometry involved)

Given isosceles triangle #AhatBC# #(AB=AC)# circumscribed in a circle with radius #r=1# .

Consider #x# the height of the triangle #AhatBC# from the vertex #A#.

  • #a)# Prove that #BC=2sqrt(x/(x-2))# #color(white)(aa)# , #x>2#

  • #b)# Find the value of #x# for which the area of the triangle #AhatBC# is minimum

  • #c)# The side #BC# of the triangle is changing with a rate of #sqrt3# #(cm)/sec#. Find the rate of change for the angle #hatA# when the triangle becomes equilateral

(Area is given as a function of #x#)

Answers

PART a):

Explanation:

Have a look:
enter image source here
I tried this:
enter image source here

PART b): (but check my maths anyway)

Explanation:

Have a look:
enter image source here

PART c) BUT I am not sure about it...I think it is wrong...

Explanation:

Have a look:
enter image source here

Part c

Explanation:

#c)#

enter image source here

Take into account that while the base #BC# of the triangle increases, the height #AM# decreases.

Based on the above,

Consider #hatA=2φ#, #color(white)(aa)# #φ##in##(0,π/2)#

We have

  • #ΔAEI#: #sinφ=1/(AI)# #<=># #AI=1/sinφ#

  • #AM=AI+IM=1/sinφ+1=(1+sinφ)/sinφ#

In #ΔAMB#: #tanφ=(MB)/(MA)# #<=># #MB=MAtanφ#

#<=># #y=(1+sinφ)/sinφ*sinφ/cosφ# #<=>#

#y=(1+sinφ)/cosφ# #<=># #y=1/cosφ+tanφ#

#<=># #y(t)=1/cos(φ(t))+tan(φ(t))#

Differentiating in respect to #t# we get

#y'(t)=(sin(φ(t))/cos^2(φ(t))+1/cos^2(φ(t)))φ'(t)#

For #t=t_0# ,

#φ=30°#

and #y'(t_0)=sqrt3/2#

Thus, since #cosφ=cos30°=sqrt3/2# and #sinφ=sin30°=1/2#

we have

#sqrt3/2=((1/2)/(3/4)+(1/3)/(3/4))φ'(t_0)# #<=>#

#sqrt3/2=(2/3+4/3)φ'(t_0)# #<=>#

#sqrt3/2=2φ'(t_0)# #<=>#

#φ'(t_0)=sqrt3/4#

But #hatA=ω(t)#, #ω(t)=2φ(t)#

therefore, #ω'(t_0)=2φ'(t_0)=2sqrt3/4=sqrt3/2# #(rad)/sec#

(Note: The moment when the triangle becomes equilateral #AI# is also the center of mass and #AM=3AI=3#, #x=3# and height= #sqrt3#)


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