Question : AB and CD are two chords in a circle with centre O and AD is a diameter. AB and CD produced meet a point P outside the circle. If $\angle A P D=25^{\circ}$ and $\angle D A P=39^{\circ}$, then the measure of $\angle C B D$ is:
Option 1: 29°
Option 2: 26°
Option 3: 27°
Option 4: 32°
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Correct Answer: 26°
Solution :
Given:
AB and CD are two chords in a circle with centre O and AD is a diameter
AB and CD produced meet at a point P outside the circle
$\angle$ APD = 25$^\circ$
$\angle$ DAP = 39$^\circ$
Concept used:
Exterior angle theorem - An exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles.
The angle subtended from a diameter is 90°
Calculation:
According to the question
We know that an exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles.
So, In $\triangle$ ADP
$\angle$ ADC = $\angle$ DAP + $\angle$ DPA
⇒ $\angle$ ADC = 25$^\circ$ + 39$^\circ$
⇒ $\angle$ ADC = 64$^\circ$
Now,
Angles in the same segment are equal
So, $\angle$ ADC = $\angle$ ABC = 64$^\circ$ ----(1)
Since AD is a diameter
So, $\angle$ ABD = 90$^\circ$
⇒ $\angle$ ABC + $\angle$ CBD = 90$^\circ$
⇒ 64$^\circ$ + $\angle$ CBD = 90$^\circ$ [From (1)]
⇒ $\angle$ CBD = 26$^\circ$
$\therefore$ The value of $\angle$CBD is 26$^\circ$.
Hence, the correct answer is 26$^\circ$.
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