Question
Download Solution PDF△PQR मध्ये, बाजू PQ व PR, E व F पर्यंत अशाप्रकारे वाढवल्या आहेत,की ∠Q व ∠R चे बाह्य कोन दुभाजक बिंदू k जवळ एकमेकांना मिळतात. जर ∠QPR = 110° असेल , तर ∠QKR = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिलेले:
∠QPR = 110°
सूत्र:
त्रिकोणाच्या तिन्ही कोनांच्या मापांची बेरीज 180° असते.
∠QKR = 90° - ∠QPR/2
पडताळा:
∠RQK = (180° - ∠PQR)/2
∠QRK = (180° - ∠PRQ)/2
△QKR मध्ये,
∠RQK + ∠QRK + ∠QKR = 180°
⇒ [(180° - ∠PQR)/2] + [(180° - ∠PRQ)/2] + ∠QKR = 180°
⇒ 90° - [∠PQR/2] + 90 – [∠PRQ/2] + ∠QKR = 180°
⇒ 180 – [(∠PQR + ∠PRQ)/2] + ∠QKR = 180
⇒ ∠QKR = (∠PQR + ∠PRQ)/2
⇒ 2∠QKR = ∠PQR + ∠PRQ
⇒ 2∠QKR = 180° - ∠QPR
⇒ 2∠QKR = 180° - 110
⇒ ∠QKR = 70/2
∴ ∠QKR = 35°
छोटीशी क्लृप्ती:
येथे, K हे ΔPQR चे बाह्यकेंद्र आहे, म्हणून ∠QKR = 90° - (∠QPR/2)
∠QKR = 90° - 110°/2
∴ ∠QKR = 90° - 55° = 35°
Last updated on Jul 10, 2025
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