Mathematics, 29.07.2021 08:00 manlyman31

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.
Need help! Please show your work too. Thanks!

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Answer from: travyse

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:

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

Let the lengths of each side of △ABC having area equal to 1 be as follows: AB = 2, BC = a and CA = b

Answer from: Quest

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6 = 0.00000180

1.80 x 10-4 = 0.000180 (bigger, or "less small", than the former one)

1.8 x 10-2 = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.

Answer from: Quest

11/36 is .305 repeating

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