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Question
In the figure, given below, it is given that AB is perpandiculer to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.
Solution
In ΔABC, we have
`tan 45^circ = (AB)/(CB) = X/(CB)` ...`[∵ tan theta = "Perpendicular"/"Base"]`
`1 = X/(CB)`
`\implies` CB = X ...(i)
Again in ΔADB, we have
`tan 30^circ = (AB)/(DB) = X/(DB)`
`\implies 1/sqrt(3) = X/(DB)`
∴ `DB = sqrt(3) xx X` ...(ii)
But DB = DC + CB
∴ `sqrt(3)X = 30 + X`
`\implies sqrt(3)X - X = 30`
`\implies X(sqrt(3) - 1) = 30`
`\implies X = 30/(sqrt(3) - 1)m`
∴ `X = (30(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`
= `(30(sqrt(3) + 1))/(3 - 1)m`
= `(30(sqrt(3) + 1))/2`
= 15 (1.732 + 1) m.
= 15 × 2.732
= 40.98 m.
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