Question
Download Solution PDFबाह्य बिंदू P पासून, केंद्र O असलेल्या वर्तुळावर स्पर्शिका PA आणि PB काढल्या जातात. जर ∠PAB= 55° असल्यास, ∠AOB शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिलेले आहे:
PA आणि PB या O केंद्र असलेल्या वर्तुळाच्या स्पर्शिका आहेत.
∠PAB = 55°
संकल्पना:
समान बाह्य बिंदूपासून काढलेल्या स्पर्शिका समान लांबीच्या असतात.
स्पर्शिकेच्या बिंदूवर स्पर्शिका त्रिज्येला लंब असतात.
गणना:
∵ ∠PAB = 55°
∴ ∠PBA = 55° (PA = PB)
त्रिकोण PAB मध्ये,
∠APB + ∠PAB + ∠PBA = 180° (त्रिकोणांच्या कोनांचा बेरजेचा गुणधर्म )
⇒ ∠P + 55° + 55° = 180°
⇒ ∠P = 70°
तसेच, ∠AOB + ∠APB = 180° (चतुर्भुजाच्या सर्व कोनांची बेरीज 360° आणि ∠P = ∠B = 90° आहे)
⇒ ∠AOB = 180° - 70° = 110
∴ ∠AOB चे माप = 110°
Last updated on Jun 27, 2025
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