Question

Let the lengths of each side of △ABC having area equal to 1 be as follows: AB = 2, BC = a and CA = b. Let CD be a perpendicular line from point C to AB. Answer the following questions.

(1) Given AD = x, write a²+(2√3-1)b² in the form of x.
(2) Find the value of x at which a²+(2√3 - 1)b² is the lowest and the magnitude of ∠BAC.
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Answer 1

Part 1)

`\displaystyle \left((2-x)^2 + 1)\right) + (2√(3) - 1 ) \left(x^2 + 1\right)`

Or simplified:

`\displaystyle = 2√(3)x^2 - 4x + 4 + 2√(3)`

Part 2)

The value of x for which the given expression will be the lowest is:

`\displaystyle x = (√(3))/(3)\approx 0.5774`

And the magnitude of ∠BAC is 60°.

Explanation:

We are given a ΔABC with an area of one. We are also given that AB = 2, BC = a, and CA = b. CD is a perpendicular line from C to AB.

Please refer to the diagram below.

Part 1)

Since we know that the area of the triangle is one:

`\displaystyle (1)/(2) (2)(CD) = 1`

Simplify:

`\displaystyle CD = 1`

From the Pythagorean Theorem:

`\displaystyle x^2 + CD^2 = b^2`

Substitute:

`x^2 + 1 = b^2`

BD will simply be (2 - x). From the Pythagorean Theorem:

`\displaystyle (2-x)^2 + CD^2 = a^2`

Substitute:

`\displaystyle (2-x)^2+ 1 = a^2`

We have the expression:

`\displaystyle a^2 + (2√(3) - 1) b^2`

Substitute:

`\displaystyle = \boxed{\left((2-x)^2 + 1)\right) + (2√(3) - 1 ) \left(x^2 + 1\right)}`

Part 2)

We can simplify the expression. Expand and distribute:

`\displaystyle (4 - 4x + x^2 + 1)+ (2√(3) -1)x^2 + 2√(3) - 1`

Simplify:

`\displaystyle = ((2√(3) -1 )x^2 + x^2) + (-4x) + (4+1-1+2√(3))`

Simplify:

`\displaystyle = 2√(3)x^2 - 4x + 4 + 2√(3)`

Since this is a quadratic with a positive leading coefficient, it will have a minimum value. Recall that the minimum value of a quadratic always occur at its vertex. The vertex is given by the formulas:

`\displaystyle \text{Vertex} = \left(-(b)/(2a), f\left(-(b)/(2a)\right)\right)`

In this case, a = 2√3, b = -4, and c = (4 + 2√3).

Therefore, the x-coordinate of the vertex is:

`\displaystyle x = -((-4))/(2(2√(3))) = (1)/(√(3)) =\boxed{ (√(3))/(3)}`

Hence, the value of x at which our expression will be the lowest is at √3/3.

To find ∠BAC, we can use the tangent ratio. Recall that:

`\displaystyle \tan \theta = \frac{\text{opposite}}{\text{adjacent}}`

Substitute:

`\displaystyle \tan \angle BAC = (CD)/(x)`

Substitute:

`\displaystyle \tan \angle BAC = (1)/((√(3))/(3)) = √(3)`

Therefore:

`\displaystyle\boxed{ m\angle BAC = \arctan√(3) = 60^\circ}`