Isosceles triangle has , and a circle with radius is tangent to line at and to line at . What is the area of the circle that passes through vertices , , and
Answers
The circle that passes through the vertices of triangle ΔABC (A, B, C) is the
circumscribing circle of triangle ΔABC.
The area of the circle that passes through vertices A, B, and C, is (C) 26·π
Reasons:
The given parameters are;
Side length of isosceles triangle ΔABC; [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex] = 3·√6
Radius of circle tangent to [tex]\overline{AB}[/tex] at B and [tex]\overline{AC}[/tex] at C = 5·√2
Required:
Area of the circle that passes through vertices A, B, and C
Solution:
Angle ∠BAO is given as follows;
[tex]\angle BAO = arctan\left(\dfrac{5 \cdot \sqrt{2} }{3 \cdot \sqrt{6}} \right) = \mathbf{arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)}[/tex]
Therefore;
[tex]\angle BOA = 90^{\circ} - arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)[/tex]
[tex]\overline{BC} = 2 \times 5 \cdot \sqrt{2} \times sin\left(90^{\circ} - arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right) = 15\cdot \sqrt{\dfrac{6}{13} }[/tex]
∠ABO' = ∠BAO' (Base angles of isosceles triangle ΔABO')
[tex]\angle BAO' = \angle BAO = \mathbf{arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)}[/tex]
Therefore;
[tex]\angle BO'A = 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)[/tex]
From sine rule, we have;
[tex]\dfrac{\overline{AB}}{sin \left(\angle BO'A \right)} = \mathbf{\dfrac{\overline{BO'}}{sin \left(\angle BAO' \right) \right)}}[/tex]
Which gives;
[tex]\mathbf{\dfrac{3 \cdot \sqrt{6} }{sin \left( 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right)}} = \dfrac{\overline{BO'}}{sin \left(arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right) \right)}[/tex]
Using a graphing calculator, we get;
[tex]\overline{BO'} = \dfrac{3 \cdot \sqrt{6} }{sin \left( 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right)} \times sin \left(arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right) \right) = \sqrt{26}[/tex]
The radius of the circumscribing circle [tex]\overline{BO'}[/tex] = √(26)
Therefore, area of the circumscribing circle, [tex]A_{O'}[/tex] = π·(√(26))² = 26·π
The area of the circle that passes through vertices A, B, and C, is (C) 26·π
Learn more here:
https://brainly.com/question/17147358
The possible question options obtained from a similar question online are;
(A) 24·π (B) 25·π (C) 26·π (D) 27·π (E) 28·π
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Answers
Answer:
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Step-by-step explanation:
Given as :
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Step-by-step explanation:
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Answers
Answer:
5%
Step-by-step explanation:
First let's find the interest,
$ 115.50 - $ 110
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Answers
Answer:
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Step-by-step explanation:
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explanation:
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Answers
Answer:
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Step-by-step explanation:
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Answer:
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Answers
Answer:
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Answers
Answer:
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Step-by-step explanation:
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Answers
Answer:
AC ≈ 50.5
Step-by-step explanation:
Using the tangent ratio in the right triangle
tan29° = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{28}{x}[/tex] ( multiply both sides by x )
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Answer:
50.5
Step-by-step explanation:
In the right triangle, focus on angle A. The side marked 28 is the opposite side, and the side marked x is the adjacent side. The trigonometric ratio that uses those two side is the tangent ratio.
tangent = (opposite) / (adjacent)
[tex]\tan{29^\circ}=\frac{28}{x}[/tex]
Multiply both sides by x.
[tex]x\cdot \tan{29^\circ}=28[/tex]
Divide by tan 29 degrees.
[tex]x=\frac{28}{\tan{29^\circ}}[/tex]
Use a calculator (in degree mode!) to approximate x.
[tex]x \approx 50.5[/tex]
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Answers
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Answer:
B) y = 2x - 3
Step-by-step explanation:
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Answer: B
Step-by-step explanation:
The slope is 2 as it shown rise (2) over run (1)
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Answer:
72.8325
Step-by-step explanation:
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Suppose an annuity pays 7.0% annual interest, compounded annually. If you invest $3,000 every year for 5 years, what percentage of the balance in the account is interest earned?
Round your answer to the nearest hundredth of a percent.
Answers
Answer:
13.05%
Step-by-step explanation:
The value of an ordinary annuity earning interest compounded annually is ...
A = P((1 +r)^n -1)/r . . . . . interest rate r for n years on principal payments P
For this account, the balance after 5 years will be ...
A = 3000((1 +.07)^5 -1)/0.07 = 3000·5.750739
The sum of payments is ...
payment total = 3000·5
Then the amount of interest is ...
3000·(5.750739 -5) = 3000·0.750739
and the fraction that is of the account balance is ...
(3000·0.750739)/(3000·5.750739) × 100% ≈ 13.05%
About 13.05% of the balance is interest earned.
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Answers
Answer: 6a3b3
Step-by-step explanation:
Reformatting the input:
Changes made to your input should not affect the solution:
(1): "0.6" was replaced by "(6/10)".
STEP 1:
Equation at the end of step 1
6
0-(((——•(a2))•b)•(0-(2•5ab2)))
10
STEP 2:
3
Simplify —
5
Equation at the end of step 2:
3
0 - (((— • a2) • b) • ( -2•5ab2))
5
STEP 3:
Equation at the end of step 3
3a2
0 - ((——— • b) • ( -2•5ab2))
5
STEP 4:
Multiplying exponential expressions :
4.1 a2 multiplied by a1 = a(2 + 1) = a3
Multiplying exponential expressions :
4.2 b1 multiplied by b2 = b(1 + 2) = b3
Canceling Out:
4.3 Canceling out 5 as it appears on both sides of the fraction line
Equation at the end of step 4:
0 - -6a3b3
STEP 5:
Final result :
6a3b3
Write the prime factorization of 185.
Answers
Answer:
5 ⋅ 37
Step-by-step explanation:
185 has factors of 5 and 37.
Answer:
Step-by-step explanation:
All Factors of 185: 1, 5, 37 and 185.
Negative Factors of 185: -1, -5, -37 and -185.
Prime Factors of 185: 5, 37.
Prime Factorization of 185: 5^1× 37^1
Sum of Factors of 185: 228.
Use the following function rule to find f(3) f(x)=3(4)^x+2
Answers
Answer:
3f×16=36
Step-by-step explanation:
f(x)=2x×fx
(substitute
=3f×16
=36f