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Home » In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB.
Circles | Class 10 | Exercise 17A | ICSE | Maths | Selina
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(A) - 3

Solution:

(i) In

    \[\vartriangle ABD,\]

    \[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\]

    \[{{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]

    \[{{135}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]

    \[\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{135}^{o}}~=\text{ }{{45}^{o}}\]

So now,

    \[\angle ADC\text{ }=\angle ADB\text{ }+\angle BDC\text{ }=\text{ }{{45}^{o}}~+\text{ }{{45}^{o}}~=\text{ }{{90}^{o}}\]

As

    \[\angle ADC\]

is the angle of semi-circle for

    \[AC\]

as the diameter of the circle.

(ii)

    \[\angle ACB\text{ }=\angle ADB\]

[Angles in the same segment of a circle]

Thus,

    \[\angle ACB\text{ }=\text{ }{{45}^{o}}\]

i

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