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CBSE Class 10 Basic Math 2025 All Sets Solved Paper

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Question : 7 of 20
Marks: +1, -0
In the given figure, AB is diameter of the circle with centre O. CD is tangent to the circle so that ∠ACD=40∘\angle ACD = 40^{\circ} The value of ∠CBA\angle CBA is
Solution:  
CD is tangent to the circle at point C.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴∠ACB=90∘\therefore \angle ACB = 90^{\circ}
∠CAB=∠ACD=40∘\angle CAB = \angle ACD = 40^{\circ}
In â–³ ACB
∠ACB+∠CBA+∠CAB=180∘\angle ACB + \angle CBA + \angle CAB = 180^{\circ}
90∘+∠CBA+40∘=180∘90^{\circ} + \angle CBA + 40^{\circ} = 180^{\circ}
130∘+∠CBA=180∘130^{\circ} + \angle CBA = 180^{\circ}
∠CBA=180∘−130∘\angle CBA = 180^{\circ} - 130^{\circ}
∠CBA=50∘\angle CBA = 50^{\circ}
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