Question
Download Solution PDF△PQRൽ, PQ, PR എന്നീ വശങ്ങൾ E, F എന്നിവയിലേക്ക് നീട്ടി, അപ്പോൾ ∠Q, ∠R എന്നിവയുടെ ബാഹ്യ കോൺ ഛേദകം k എന്ന പോയിന്റിൽ കണ്ടുമുട്ടുന്നു. QPR = 110° ആണെങ്കിൽ, ∠QKR =?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFതന്നിരിക്കുന്നത്:
∠QPR = 110°
സമവാക്യം:
ഒരു ത്രികോണത്തിന്റെ മൂന്ന് കോണുകളുടെ ആകെത്തുക 180° ആണ്.
∠QKR = 90° - ∠QPR/2
കണക്കുകൂട്ടൽ:
∠RQK = (180° - ∠PQR)/2
∠QRK = (180° - ∠PRQ)/2
△QKRൽ
∠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°
എളുപ്പവഴി:
ഇവിടെ, K ആണ് ΔPQRന്റെ എക്സ് കേന്ദ്രം, അതിനാൽ ∠QKR = 90° - (∠QPR/2)
∠QKR = 90° - 110°/2
∴ ∠QKR = 90° - 55° = 35°
Last updated on Jul 10, 2025
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