Question
A Navy captain going away from a lighthouse at the speed
of 4[(√3) – 1] m/s. He observes that it takes him 1 minute to change the angle of elevation of the top of the lighthouse from 60o to 45o. What is the height (in metres) of the lighthouse?Solution
Let height is H tan 600 = H / BD = BD = H / tan600 [BD = H / √3 tan 450 = AB / BC = AB = BC, BC = H 4(√3 – 1) m/s speed Time = 1 minute 60 Second Distance = Speed × Time = 4√3 – 1) × 60 = 240√3 – 240 CD = 240√3 – 60 CD = H – H/√3 = 240√3 – 240 H(√3 – 1) = 240√3(√3 – 1) H = 240√3 m
?% of 1499.89 + 54.14 Γ 8 = 25.05% of 5568.08
? = 41.92% of (34.92 x 40.42) + 29.78% of 399.84
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
(15.87% of 79.98 + 19.69% of 64.22) Γ 4.83 = ?
? + 163.99 β 108.01 = 25.01 Γ 6.98
80.09 * β144.05+ ? * β224.87 = (2109.09 Γ· β1368.79) * 19.89
(15.15Β Γ Β 31.98) + 30.15% of 719.99 = ? + 124.34
(124.99)Β² = ?
6.992 + (2.01 Γ 2.98) + ? = 175.03
(627.98 Γ· 3.98 + 11.01 X 12.98 - ?) Γ· β623 = (178.98 + 37.08) Γ· 23.98Β