Math, asked by seemforsun, 10 months ago

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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Answered by amitnrw
14

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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Answered by swarshah2217
0

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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