Question
Download Solution PDFकेंद्र O वाले एक वृत्त पर एक बाह्य बिंदु P से, स्पर्शरेखा PA और PB खींची जाती है। यदि ∠PAB= 55° है, तो ∠AOB ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
PA और PB केंद्र O वाले वृत्त की स्पर्श रेखाएँ हैं।
∠PAB= 55°
अवधारणा:
एक ही बाह्य बिंदु से खींची गई स्पर्श रेखाएं लंबाई में बराबर होती हैं।
स्पर्शरेखा के बिंदु पर त्रिज्या के लंबवत स्पर्शरेखा।
गणना:
∵ ∠PAB = 55°
∴ ∠PBA = 55° (PA = PB)
त्रिभुज PAB में,
∠APB + ∠PAB + ∠PBA = 180° (त्रिभुज का कोण योग गुण)
⇒ ∠P + 55° + 55° = 180°
⇒ ∠P = 70°
साथ ही, ∠AOB + ∠APB = 180° (एक चतुर्भुज के सभी कोणों का योग 360° और ∠P = 70 ° ∠B = 90° है)
⇒ ∠AOB = 180° - 70° = 110
∴ ∠AOB की माप = 110°
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