Question
Download Solution PDFAB, CD के समांतर है। एक तिर्यक रेखा PQ, AB और CD को क्रमशः E और F पर प्रतिच्छेद करती है, और ∠PEB = 56° है। AB और CD के बीच एक बिंदु G इस प्रकार है कि ∠BEG = 32° और ∠GFE = 47° है। ∠EGF का मान कितना है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
AB, CD के समांतर है।
तिर्यक रेखा PQ, AB और CD को क्रमशः E और F पर प्रतिच्छेद करती है।
∠PEB = 56°
∠BEG = 32°
∠GFE = 47°
प्रयुक्त सूत्र:
किसी त्रिभुज में, कोणों का योग 180° होता है:
∠GEF + ∠GFE + ∠EGF = 180°
गणना:
∠BEG = 32°
∠GFE = 47°
⇒ ∠GEF = 180° - (∠BEG + ∠PEB)
⇒ ∠GEF = 180° - (32° + 56°)
⇒ ∠GEF = 180° - 88°
⇒ ∠GEF = 92°
अब, ∠GEF + ∠GFE + ∠EGF = 180°
⇒ 92° + 47° + ∠EGF = 180°
⇒ 139° + ∠EGF = 180°
⇒ ∠EGF = 180° - 139°
⇒ ∠EGF = 41°
इसलिए, सही उत्तर विकल्प (3) है।
Last updated on Jul 17, 2025
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