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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Find the size of the angles marked by letters in the following diagram.
Answers
Answer:
x = 52
y = 19
Step-by-step explanation:
The degree measure of a straight line is (180) degrees. One can use this to find the measure of the angle that is adjacent to the angle with a measure of (142) degrees. Form an equation and solve for the unknown,
142 + (unknown angle) = 180
unknown angle = 38
The inscribed angle theorem states that twice the measure of an angle with its vertex of the circumference (outer edge) of a circle is equal to the measure of the surrounding arc. Therefore, one can state the following:
(2)(unknown angle) = (surrounding arc)
(2)(38) = (surrounding arc)
76 = surrounding arc
The central angles theorem states that the measure of an angle whose vertex is the center of the circle is equal to the measure of the surrounding arc. Therefore, one can state the following:
m<O = (surrounding arc)
m<O = 76
The radius is the distance from the center of the circle to the circumference of the circle. All radii in a circle are congrunet. Therefore the triangle with angle (x) and vertex (O) is an isosceles triangle, as two of its sides are radii, and are thus congruent. One property of an isosceles triangle is the base angles theorem. This theorem states that the angles opposite the congruent sides in an isosceles triangle are congruent. Moreover, the sum of angles in any triangle is (180) degrees. Therefore, one can make the following statement:
m<O + x + x = 180
76 + 2x = 180
x = 52
Finally, one can use the property that the sum of angles in any triangle is (180) degrees to make the following statement:
(x + 19) + (x + y) + (38) = 180
52 + 19 + 52 + y + 38 = 180
161 + y = 180
y = 19
x=52°
y=19°
Answer:
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°
x=104/2
x=52°
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O x=-3
O y = -3
O x + y =-3
O X-y=-3
Answers
Answer:
Option A, x = -3
Step-by-step explanation:
Step 1: Find the graphed line
x = -3
Answer: Option A, x = -3
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Step-by-step explanation:
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Answer:
y = -x/2 - 14
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Drag the tiles to the correct boxes to complete the pairs.
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three vertices of square ABCD. What are
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three vertices of square ABCD. What are
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3.75 < n
Step-by-step explanation:
n + 7 < 5n - 8
Add 8 to both sides
n + 7 + 8 < 5n
n + 15 < 5n
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15 < 4n
Divide both sides by 4
[tex]\frac{15}{4}[/tex] < n
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Answer:
Step-by-step explanation:
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Squaring this ensures us that we don't end up with zero, which would be useless.
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Answers
Step-by-step explanation:
Answer with explanation is in the attachment
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Step-by-step explanation:
the triangle inequality theorem states that any 2 sides of a triangle must be greater than the measure of the third side. now with this rule, let's apply it to all options!
option a:
7 + 5 > 6 .. yes!
6+ 7 > 5 .. yes!
5 + 6 > 7 .. yes!
option b:
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12 + 19 > 9 .. no!
19 + 9 > 12 .. no!
option c:
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11 + 18 > 7 .. no!
18 + 7 > 11 .. yes!
option d:
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8 + 11 > 6 .. yes!
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Given:
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Step-by-step explanation:
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Answers
Answer:
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uehdjd
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Answers
Answer:
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Answers
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Answers
Answer:
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Step-by-step explanation:
