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If O is circumcentre of acute angled triangle ABC, then ∠ BOC = 2 ∠ BAC 100 = 2 ∠ BAC ∠ BAC= 500
(√4623.9 + √484.2) – √2303.97 ÷ √1296.4 × √35.98 ÷ √15.99 = ?
181.87 ÷ 13.89 X 8.13 + ? = 11.852
(27.08)2 – (14.89)2 – (22.17)2 = ?
509.85 ÷ 15.05 + 210.16 – 18.06 × 5.95 = ?
89.87 × 3.21 + 60.32 = ? × (6.89 2 – 19.21)
12.5% of 6400 + (17 × 25) = ?% of 2200+ 125