AB is a diameter of a circle with centre O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P. If ∠ APC = 62°, then find the measure of the minor arc AC. (i.e.∠ ABC).
Minor arc AC will create angle CBA ∠APC = 62º = ∠APB ∠BAP = 90° (diameter perpendicular to tangent) In Δ APB, ∠APB + ∠BAP + ∠PBA = 180° ⇒ ∠PBA = 180° - (62° + 62°) ⇒ ∠PBA = 56° ∴ The measure of minor arc AC =56°
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)
63- [22-{24 ÷ 3-(9-15 ÷ 5) ÷ 6}]=?
1000÷ 250 = ( 3√? × √1444) ÷ ( 3√512 × √361)
(〖(0.4)〗^(1/3) × 〖(1/64)〗^(1/4) × 〖16〗^(1/6) × 〖(0.256)〗^(2/3))/(〖(0.16)〗^(2/3) × 4^(-1/2) ×〖1024〗^(-1/4) ) = ?
(3500 ÷ √1225) × √(20.25) = ? ÷ 4
√ 225 x 24 - √ 144 x 18 = ?
[123 ÷ 8 ÷ 9] × 144 = ? + 12 × 5
[(36 × 15 ÷ 96 + 19 ÷ 8) × 38] = ?% of 608
2/5 of 3/4 of 7/9 of 14400 = ?
(√196 + √121) × 4 = ?/2
