in the figure, points P B and Q are points of contacts of the respective tangents, line QA is parallel to the line PC . If QA=7.2cm , PC=5cm , find the radius.
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Answers
Radius = 6 cm
Step-by-step explanation:
∠OQA = 90° ∠OBA = 90° ∠OPC = 90° ∠OBC = 90°
Using sum of angles of Quadrilateral = 360°
Let say ∠QAB = α => ∠QOB = 180-α
=> ∠PCB = 180 - α and ∠POB = α
OQ = OB = OP = Radius
QB = QA = 7.2 cm ( Equal Tangent)
BC = PC = 5.2 cm (Equal Tangent)
Using Cosine formula of triangle
QB² = 7.2² + 7.2² - 2*7.2²Cosα
QB² = R² + R² - 2*R²Cos(180 - α)
=> 7.2² + 7.2² - 2*7.2²Cosα = R² + R² + 2*R²Cosα
=> 7.2² * 2 ( 1 - Cosα) = 2 R² (1 + Cosα)
=> ( 1 - Cosα)/ (1 + Cosα) = R²/7.2²
PB² = R² + R² - 2*R²Cosα
PB² = 5² + 5² - 2*5²Cos(180 - α)
=> 2R²(1 - Cosα) = 2*5²(1 + Cosα)
=> ( 1 - Cosα)/ (1 + Cosα) = 5²/R²
R²/7.2² = 5²/R²
=> R² * R² = 7.2² * 5²
=> R * R = 7.2 * 5
=> R² = 36
=> R = 6
Radius = 6 cm
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Radius = 6 cm
Step-by-step explanation:
∠OQA = 90° ∠OBA = 90° ∠OPC = 90° ∠OBC = 90°
Using sum of angles of Quadrilateral = 360°
Let say ∠QAB = α => ∠QOB = 180-α
=> ∠PCB = 180 - α and ∠POB = α
OQ = OB = OP = Radius
QB = QA = 7.2 cm ( Equal Tangent)
BC = PC = 5.2 cm (Equal Tangent)
Using Cosine formula of triangle
QB² = 7.2² + 7.2² - 2*7.2²Cosα
QB² = R² + R² - 2*R²Cos(180 - α)
=> 7.2² + 7.2² - 2*7.2²Cosα = R² + R² + 2*R²Cosα
=> 7.2² * 2 ( 1 - Cosα) = 2 R² (1 + Cosα)
=> ( 1 - Cosα)/ (1 + Cosα) = R²/7.2²
PB² = R² + R² - 2*R²Cosα
PB² = 5² + 5² - 2*5²Cos(180 - α)
=> 2R²(1 - Cosα) = 2*5²(1 + Cosα)
=> ( 1 - Cosα)/ (1 + Cosα) = 5²/R²
R²/7.2² = 5²/R²
=> R² * R² = 7.2² * 5²
=> R * R = 7.2 * 5
=> R² = 36
=> R = 6
Radius = 6 cm