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Standard VII

Mathematics

Angle Sum Property

The diagram s...

Question

The diagram shows three touching circles.
$A$ is the centre of a circle of radius $r$ centimeter.
$B$ and $C$ are the centre of circle of radius $3.8$ centimeters. Angle $A B C = 70^{o}$ .
Find the $∠ A$

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Solution

Let the point of intersection of circles $A, B$ be $P$
Let the point of intersection of circles $A, C$ be $Q$
$A P = A Q$
$B P = C Q$
$⟹ A B + A C ⟹ ∠ B = ∠ C$
By angle sum property
$∠ A + ∠ B + ∠ C = 180 ∠ A = 70 + 70 = 180 ∠ A = 180 - 140 = 40$

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The diagram shows three touching circles.A is the centre of a circle of radius r centimeter.B and C are the centre of circle of radius 3.8 centimeters. Angle ABC=70o.Find the ∠A

Let the point of intersection of circles A,B be PLet the point of intersection of circles A,C be QAP=AQ BP=CQ⟹AB+AC⟹∠B=∠CBy angle sum property ∠A+∠B+∠C=180∠A=70+70=180∠A=180−140=40

The diagram shows three touching circles.
$A$ is the centre of a circle of radius $r$ centimeter.
$B$ and $C$ are the centre of circle of radius $3.8$ centimeters. Angle $A B C = 70^{o}$ .
Find the $∠ A$

Let the point of intersection of circles $A, B$ be $P$
Let the point of intersection of circles $A, C$ be $Q$
$A P = A Q$
$B P = C Q$
$⟹ A B + A C ⟹ ∠ B = ∠ C$
By angle sum property
$∠ A + ∠ B + ∠ C = 180 ∠ A = 70 + 70 = 180 ∠ A = 180 - 140 = 40$