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!
Answers
Answer:
Part 1)
[tex]\displaystyle \left((2-x)^2 + 1)\right) + (2\sqrt{3} - 1 ) \left(x^2 + 1\right)[/tex]
Or simplified:
[tex]\displaystyle = 2\sqrt{3}x^2 - 4x + 4 + 2\sqrt{3}[/tex]
Part 2)
The value of x for which the given expression will be the lowest is:
[tex]\displaystyle x = \frac{\sqrt{3}}{3}\approx 0.5774[/tex]
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:
[tex]\displaystyle \frac{1}{2} (2)(CD) = 1[/tex]
Simplify:
[tex]\displaystyle CD = 1[/tex]
From the Pythagorean Theorem:
[tex]\displaystyle x^2 + CD^2 = b^2[/tex]
Substitute:
[tex]x^2 + 1 = b^2[/tex]
BD will simply be (2 - x). From the Pythagorean Theorem:
[tex]\displaystyle (2-x)^2 + CD^2 = a^2[/tex]
Substitute:
[tex]\displaystyle (2-x)^2+ 1 = a^2[/tex]
We have the expression:
[tex]\displaystyle a^2 + (2\sqrt{3} - 1) b^2[/tex]
Substitute:
[tex]\displaystyle = \boxed{\left((2-x)^2 + 1)\right) + (2\sqrt{3} - 1 ) \left(x^2 + 1\right)}[/tex]
Part 2)
We can simplify the expression. Expand and distribute:
[tex]\displaystyle (4 - 4x + x^2 + 1)+ (2\sqrt{3} -1)x^2 + 2\sqrt{3} - 1[/tex]
Simplify:
[tex]\displaystyle = ((2\sqrt{3} -1 )x^2 + x^2) + (-4x) + (4+1-1+2\sqrt{3})[/tex]
Simplify:
[tex]\displaystyle = 2\sqrt{3}x^2 - 4x + 4 + 2\sqrt{3}[/tex]
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:
[tex]\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)[/tex]
In this case, a = 2√3, b = -4, and c = (4 + 2√3).
Therefore, the x-coordinate of the vertex is:
[tex]\displaystyle x = -\frac{(-4)}{2(2\sqrt{3})} = \frac{1}{\sqrt{3}} =\boxed{ \frac{\sqrt{3}}{3}}[/tex]
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:
[tex]\displaystyle \tan \theta = \frac{\text{opposite}}{\text{adjacent}}[/tex]
Substitute:
[tex]\displaystyle \tan \angle BAC = \frac{CD}{x}[/tex]
Substitute:
[tex]\displaystyle \tan \angle BAC = \frac{1}{\dfrac{\sqrt{3}}{3}} = \sqrt{3}[/tex]
Therefore:
[tex]\displaystyle\boxed{ m\angle BAC = \arctan\sqrt{3} = 60^\circ}[/tex]
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Solution given:
<OCA=19°[base angle of isosceles triangle]
Since ∆AOC is similar to ∆ BOC
y=19°[corresponding angle of a similar triangle are equal]
<OAB=<OBA=x[base angle of isosceles triangle]
again
<A+<B=180°[exterior angle of a triangle is equal to the sum of two opposite interior angle]
x+19+x+19=142°
2x=142°-38°
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uehdjd
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Answers
Answer:
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Answers
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Please HELP! ):
The measure of an interior angle of a regular polygon is eight times the measure of one of its exterior angles. How many sides does the polygon have?
