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In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16?
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In a triangle ABC angles are in arithmetic progression with common dif...
Given:
- Triangle ABC with angles in arithmetic progression with common difference α.
- cos(α) = 21/22.
- The triangle is inscribed in a circle with radius 11 units.
- H is the orthocenter of the triangle.
To find:
The value of HA * HB * HC / 16.
Solution:
Step 1: Determine the measure of each angle in the triangle.
Let the angles of the triangle be A, A + α, and A + 2α.
Using the property that the sum of the angles in a triangle is 180 degrees, we have:
A + A + α + A + 2α = 180
3A + 3α = 180
A + α = 60
Since the angles are in an arithmetic progression, we have:
A + 2α - (A + α) = α = 60 - A
Step 2: Determine the values of A and α.
Using the given value of cos(α) = 21/22, we can find the value of sin(α) using the Pythagorean identity:
sin^2(α) + cos^2(α) = 1
sin^2(α) = 1 - cos^2(α)
sin^2(α) = 1 - (21/22)^2
sin^2(α) = 484/484 - 441/484
sin^2(α) = 43/484
sin(α) = √(43/484)
sin(α) = √43 / 22
We know that sin(α) = sin(π/2 - α). Therefore, sin(π/2 - α) = √43 / 22.
Using the identity sin(π/2 - α) = √(1 - cos^2(α)), we have:
√(1 - cos^2(α)) = √43 / 22
1 - cos^2(α) = 43 / 484
cos^2(α) = 441 / 484
cos(α) = ± 21 / 22
Since α is acute, we take the positive value: cos(α) = 21 / 22.
Now, we have cos(α) = 21 / 22, and α = 60 - A.
Substituting these values into the equation cos(α) = 21 / 22, we get:
cos(60 - A) = 21 / 22
Using the property cos(60 - A) = cos(A - 60), we can rewrite the equation as:
cos(A - 60) = 21 / 22
Step 3: Solve for A.
To solve this equation, we can use the inverse cosine function:
A - 60 = arccos(21 / 22)
A = arccos(21 / 22) + 60
Using a calculator, we find A ≈ 18.02 degrees.
Step 4: Calculate the measures of each angle.
Since A + α = 60, we have:
18.02 + α = 60
α = 60 - 18.02
- Triangle ABC with angles in arithmetic progression with common difference α.
- cos(α) = 21/22.
- The triangle is inscribed in a circle with radius 11 units.
- H is the orthocenter of the triangle.
To find:
The value of HA * HB * HC / 16.
Solution:
Step 1: Determine the measure of each angle in the triangle.
Let the angles of the triangle be A, A + α, and A + 2α.
Using the property that the sum of the angles in a triangle is 180 degrees, we have:
A + A + α + A + 2α = 180
3A + 3α = 180
A + α = 60
Since the angles are in an arithmetic progression, we have:
A + 2α - (A + α) = α = 60 - A
Step 2: Determine the values of A and α.
Using the given value of cos(α) = 21/22, we can find the value of sin(α) using the Pythagorean identity:
sin^2(α) + cos^2(α) = 1
sin^2(α) = 1 - cos^2(α)
sin^2(α) = 1 - (21/22)^2
sin^2(α) = 484/484 - 441/484
sin^2(α) = 43/484
sin(α) = √(43/484)
sin(α) = √43 / 22
We know that sin(α) = sin(π/2 - α). Therefore, sin(π/2 - α) = √43 / 22.
Using the identity sin(π/2 - α) = √(1 - cos^2(α)), we have:
√(1 - cos^2(α)) = √43 / 22
1 - cos^2(α) = 43 / 484
cos^2(α) = 441 / 484
cos(α) = ± 21 / 22
Since α is acute, we take the positive value: cos(α) = 21 / 22.
Now, we have cos(α) = 21 / 22, and α = 60 - A.
Substituting these values into the equation cos(α) = 21 / 22, we get:
cos(60 - A) = 21 / 22
Using the property cos(60 - A) = cos(A - 60), we can rewrite the equation as:
cos(A - 60) = 21 / 22
Step 3: Solve for A.
To solve this equation, we can use the inverse cosine function:
A - 60 = arccos(21 / 22)
A = arccos(21 / 22) + 60
Using a calculator, we find A ≈ 18.02 degrees.
Step 4: Calculate the measures of each angle.
Since A + α = 60, we have:
18.02 + α = 60
α = 60 - 18.02
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In a triangle ABC angles are in arithmetic progression with common dif...
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In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16?.
In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a triangle ABC angles are in arithmetic progression with common difference alpha such that cos( alpha) is equal to 21/22 and the triangle is inscribed in a circle with radius 11 units and H is orthocentre of the triangle then find HA HB HC / 16?.
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