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!

Part 1)

Or simplified:

Part 2)

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

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:

Simplify:

From the Pythagorean Theorem:

Substitute:

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

Substitute:  

We have the expression:

Substitute:

Part 2)

We can simplify the expression. Expand and distribute:

Simplify:

Simplify:

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:

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

Therefore, the x-coordinate of the vertex is:

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:

Substitute:

Substitute:

Therefore:

d. 9

step-by-step explanation:

54/12 = 4.5

sf = 4.5

so x = 2 · 4.5 = 9

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3