In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70degree and angle DBC = 30degree, calculate angle BDC. - Noon Academy

In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70degree and angle DBC = 30degree, calculate angle BDC.

Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords

In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70degree and angle DBC = 30degree, calculate angle BDC.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 7

Solution:

$ABCD$

is a cyclic quadrilateral,

$\angle BCD\text{ }+~\angle BAD\text{ }=\text{ }{{180}^{\circ }}$

[Opposite angles of a cyclic quadrilateral are supplementary]

$\angle BCD\text{ }+\text{ }{{70}^{o}}~=\text{ }{{180}^{\circ }}$

$\angle BCD\text{ }=\text{ }{{180}^{o}}~\text{ }{{70}^{o}}~=\text{ }{{110}^{o}}$

And, by angle sum property of

$\vartriangle BCD$

we have

$\angle CBD\text{ }+~\angle BCD\text{ }+~\angle BDC\text{ }=\text{ }{{180}^{o}}$

${{30}^{o}}~+\text{ }{{110}^{o}}~+~\angle BDC\text{ }=\text{ }{{180}^{o}}$

Or,

$\angle BDC\text{ }=\text{ }{{180}^{o~}}-\text{ }{{140}^{o}}$

Thus,

$\angle BDC\text{ }=\text{ }{{40}^{o}}$

i

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