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In the figure ∠BAD = 3 ∠DBA, find ∠CDB. ∠DBC and ∠ABC. |
Answer» In ΔBCD, ∠DBC + ∠BCD = ∠BDA (exterior angle property) ∠DBC + 65° -104° ∴ ∠DBC = 104° – 65° = 39° Also ∠BDA + ∠BDC = 180° (linear pair of angles) 104° + ∠BDC = 180° ∠CDB or ∠BDC = 180° – 104° = 76° Now in ΔABD, ∠BAD + ∠ADB + ∠DHA = 180° 3∠DBA + ∠DBA + 104° = 180° (given) 4∠DBA + 104° – 180° ∴ 4∠DBA = 180° – 104°= 76° ∴∠DBA = 76°/4 = 19° Now ∠ABC = ∠DBA + ∠DBC = 19° + 39° = 58° |
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