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. |
Let ∠ACB = Θ.
Then, | AC | = | √3 ⟹ cot Θ = √3 |
AB |
∴ Θ = 30°.
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. |
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.
Rate = | ❨ | 100 x 3000 | ❩% | = 6% |
12500 x 4 |
Then, | ❨ | 1200 x R x R | ❩ | = 432 |
100 |
⟹ 12R2 = 432
⟹ R2 = 36
⟹ R = 6.
Time = | ❨ | 100 x 81 | ❩years | = 4 years. |
450 x 4.5 |
Principal |
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= Rs. 8925. |
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
∴ Principal = Rs. (815 - 117) = Rs. 698.
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