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

Answer:

Part 1)

$\displaystyle \left((2-x)^2 + 1)\right) + (2\sqrt{3} - 1 ) \left(x^2 + 1\right)$

Or simplified:

$\displaystyle = 2\sqrt{3}x^2 - 4x + 4 + 2\sqrt{3}$

Part 2)

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

$\displaystyle x = \frac{\sqrt{3}}{3}\approx 0.5774$

And the magnitude of ∠BAC is 60°.

Step-by-step 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:

$\displaystyle \frac{1}{2} (2)(CD) = 1$

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:

$\displaystyle a^2 + (2\sqrt{3} - 1) b^2$

Substitute:

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

Part 2)

We can simplify the expression. Expand and distribute:

$\displaystyle (4 - 4x + x^2 + 1)+ (2\sqrt{3} -1)x^2 + 2\sqrt{3} - 1$

Simplify:

$\displaystyle = ((2\sqrt{3} -1 )x^2 + x^2) + (-4x) + (4+1-1+2\sqrt{3})$

Simplify:

$\displaystyle = 2\sqrt{3}x^2 - 4x + 4 + 2\sqrt{3}$

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:

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

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

Therefore, the x-coordinate of the vertex is:

$\displaystyle x = -\frac{(-4)}{2(2\sqrt{3})} = \frac{1}{\sqrt{3}} =\boxed{ \frac{\sqrt{3}}{3}}$

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:

$\displaystyle \tan \angle BAC = \frac{CD}{x}$

Substitute:

$\displaystyle \tan \angle BAC = \frac{1}{\dfrac{\sqrt{3}}{3}} = \sqrt{3}$

Therefore:

$\displaystyle\boxed{ m\angle BAC = \arctan\sqrt{3} = 60^\circ}$