Question

AP and BP are tangents to the circle with centre 0. If angle CBP = 25° and angle CAP =40 .Find

1) angle ADB
ii) angle AOB iii) angle ACB iv) angle APB
​ plz answer fast...its urgent

Answer 1

(i) ∠ADB = 65°

(ii) ∠AOB = 130°

(iii) ∠ACB = 115°

(iv) ∠APB = 50°

Solution:

Given AP and BP are tangents to the circle with center O.

∠CBP = 25° and ∠CAP = 40°

(i) Angles in the alternate segments are equal.

$\angle \mathrm{CDB}=\angle \mathrm{CBP}$

$\therefore \angle C D B=25^{\circ}$ – – – – (1)

Similarly, Angles in the alternate segments are equal.

$\angle \mathrm{CDA}=\angle \mathrm{CAP}=40^{\circ}$

$\therefore \angle \mathrm{ADB}=\angle \mathrm{CDA}+\angle \mathrm{CDB}$

              $=40^{\circ}+25^{\circ}$

              $=65^{\circ}$

Therefore, ∠ADB = 65°.

(ii) Arc ACB subtends ∠AOB at the center and ∠ADB at the remaining part of the circle.

$\therefore \angle A O B=2 \angle A D B$

               $=2 \times 65^{\circ}$

               $=130^{\circ}$

Therefore, ∠AOB = 130°.

(iii) ACBD is a cyclic quadrilateral.

$\therefore \angle \mathrm{ACB}+\angle \mathrm{ADB}=180^{\circ}$

$\Rightarrow \angle A C B+65^{\circ}=180^{\circ}$

$\Rightarrow \angle A C B=180^{\circ}-65^{\circ}$

               $=115^{\circ}$

Therefore, ∠ACB = 115°.

(iv) $\angle \mathrm{AOB}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow 130^{\circ}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow \angle \mathrm{APB}=180^{\circ}-130^{\circ}$

$\Rightarrow \angle \mathrm{APB}=50^{\circ}$

Therefore, ∠APB = 50°.

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