tan Θ = | AB |
BQ |
tan Θ = | h | ...(i) |
b |
tan (90° – Θ) = | AB |
BP |
cot Θ = | h | ...(ii) [∵ tan (90° – Θ) = cot Θ] |
a |
tan Θ cot Θ = | h | × | h |
b | a |
1 = | h2 | [∵ tan Θ cot Θ = 1] |
ab |
h = | ab |
tan 30° = | AB |
BC |
1 | = | h |
3 | 100 |
h = | 100 | m |
3 |
tan Θ = | AB | = | h | ...(i) |
BC | 100 |
tan 2Θ = | AB |
BD |
2tanΘ | = | h | [ | ∵ tan 2Θ = | 2tanΘ | ] |
1 – tan2 Θ | 60 | 1 – tan2 Θ |
120 × | h | = | h | [ | 1 – | ( | h | ) | 2 | ] | [From eq. (i)] |
160 | 160 |
3 | = 1 – | ( | h | ) | 2 |
4 | 160 |
( | h | ) | 2 | = 1 – | 3 | = | 1 |
160 | 4 | 4 |
h | = |
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= | 1 | ||
160 | 2 |
tan 30° = | AB |
BC |
1 | = | h |
3 | 50 |
h = | 50 | m |
3 |
tan 45° = | AB | ⇒ |
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BD |
tan 30° = | AB | ⇒ |
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BC |
⇒ h | 3 | – x = 10 |
⇒ h | 3 | – h = 10 [From eq. (i)] |
⇒ h( | 3 | – 1) = 10 |
⇒ h = | 10 | × |
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= 5( |
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+ 1) m | |||
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