Question
Download Solution PDFএকটি বাহ্যিক বিন্দু P থেকে, কেন্দ্র O সহ একটি বৃত্তে স্পর্শক PA এবং PB আঁকা হয়। ∠PAB= 55° হলে, ∠AOB এর মান নির্ণয় করুন।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFপ্রদত্ত:
PA এবং PB হল O কেন্দ্রবিশিষ্ট বৃত্তের স্পর্শক।
∠PAB= 55°
ধারণা:
একই বাহ্যিক বিন্দু থেকে আঁকা স্পর্শকগুলির দৈর্ঘ্য সমান হয়।
একটি স্পর্শক, স্পর্শক বিন্দুতে ব্যাসার্ধের ওপর লম্ব হয়।
গণনা:
∵ ∠PAB = 55°
∴ ∠PBA = 55° (PA = PB)
PAB ত্রিভুজে,
∠APB + ∠PAB + ∠PBA = 180° (কোণ সমষ্টি বৈশিষ্ট্য)
⇒ ∠P + 55° + 55° = 180°
⇒ ∠P = 70°
এছাড়াও, ∠AOB + ∠APB = 180° (একটি চতুর্ভুজের সমস্ত কোণের সমষ্টি হল 360° & ∠P = ∠B = 90°)
⇒ ∠AOB = 180° - 70° = 110
∴ ∠AOB এর মান হল = 110°
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