Draw BE ⊥ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20√3 m.
DE | = tan 30° = | 1 |
BE | √3 |
⟹ DE = | 20√3 | m = 20 m. |
√3 |
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.
Then, ∠ACB = 60° and AC = 4.6 m.
AC | = cos 60° = | 1 |
BC | 2 |
⟹ BC | = 2 x AC |
= (2 x 4.6) m | |
= 9.2 m. |
So, the data is inadequate.
4.5 km/hr = | ❨ | 4.5 x | 5 | ❩ | m/sec = | 5 | m/sec = 1.25 m/sec, and |
18 | 4 |
5.4 km/hr = | ❨ | 5.4 x | 5 | ❩ | m/sec = | 3 | m/sec = 1.5 m/sec. |
18 | 2 |
Let the speed of the train be x m/sec.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
⟹ 8.4x - 10.5 = 8.5x - 12.75
⟹ 0.1x = 2.25
⟹ x = 22.5
∴ Speed of the train = | ❨ | 22.5 x | 18 | ❩ | km/hr = 81 km/hr. |
5 |
Relative speed | = (x + 50) km/hr | |||||||
|
||||||||
|
Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
|
⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
∴ | (100 + 100) | = 3x |
8 |
⟹ 24x = 200
⟹ x = | 25 | . |
3 |
So, speed of the faster train = | 50 | m/sec |
3 |
= | ❨ | 50 | x | 18 | ❩km/hr |
3 | 5 |
= 60 km/hr.
Then, ∠APB = 30° and AB = 100 m.
AB | = tan 30° = | 1 |
AP | √3 |
⟹ AP | = (AB x √3) m |
= 100√3 m | |
= (100 x 1.73) m | |
= 173 m. |
Let ∠ACB = Θ.
Then, | AC | = | √3 ⟹ cot Θ = √3 |
AB |
∴ Θ = 30°.
Then, AB = 100 m, ∠ACB = 30° and ∠ADB = 45°.
AB | = tan 30° = | 1 | ⟹ AC = AB x √3 = 100√3 m. |
AC | √3 |
AB | = tan 45° = 1 ⟹ AD = AB = 100 m. |
AD |
∴ CD = (AC + AD) | = (100√3 + 100) m |
= 100(√3 + 1) | |
= (100 x 2.73) m | |
= 273 m. |
Rate = | ❨ | 100 x 3000 | ❩% | = 6% |
12500 x 4 |
Then, | ❨ | 1200 x R x R | ❩ | = 432 |
100 |
⟹ 12R2 = 432
⟹ R2 = 36
⟹ R = 6.
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