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`)

Answer 1

PART a):

Explanation:

Have a look:

I tried this:

Answer 2

PART b): (but check my maths anyway)

Explanation:

Have a look:

Answer 3

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

Explanation:

Have a look:

Answer 4

Part c

Explanation:

`c)`

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`)