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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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:

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

ans789

step-by-step explanation: