In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65o, calculate ∠DEC. - Noon Academy

In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65o, calculate ∠DEC.

Circles | Class 10 | Exercise 17C | ICSE | Maths | Selina

In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65o, calculate ∠DEC.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 5

Solution:

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 6

Join

$OE$

$Arc\text{ }EC$

subtends

$\angle EOC$

at the centre and

$\angle EBC$

at the remaining part of the circle.

$\angle EOC\text{ }=\text{ }2\angle EBC\text{ }=\text{ }2\text{ }x\text{ }{{65}^{o}}~=\text{ }{{130}^{o}}$

Now, in

$\vartriangle OEC$

$OE\text{ }=\text{ }OC$

[Radii of the same circle]

So,

$\angle OEC\text{ }=\angle OCE$

But, in

$\vartriangle EOC$

by angle sum property

$\angle OEC\text{ }+\angle OCE\text{ }+\angle EOC\text{ }=\text{ }{{180}^{o}}$

[Angles of a triangle]

$\angle OCE\text{ }+\angle OCE\text{ }+\angle EOC\text{ }=\text{ }{{180}^{o}}$

Or,

$2\angle OCE\text{ }+\text{ }{{130}^{o}}~=\text{ }{{180}^{o}}$

$2\angle OCE\text{ }=\text{ }{{180}^{o}}-\text{ }{{130}^{o}}$

So,

$\angle OCE\text{ }=\text{ }{{50}^{o}}/\text{ }2\text{ }=\text{ }{{25}^{o}}$

And,

$AC\text{ }||\text{ }ED$

[Given]

$\angle DEC\text{ }=\angle OCE$

[Alternate angles]

Thus,

$\angle DEC\text{ }=\text{ }{{25}^{o}}$

i

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