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Question
Find x and y, in each of the following figure:
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Solution
In right ΔABC,
tan45° = `"AB"/"BC"`
⇒ 1 = `x/(15 + y)`
⇒ x = 15 + y ....(i)
In right ΔABD,
tan60° = `"AB"/"BD"`
⇒ `sqrt(3) = x/y`
⇒ `sqrt(3) = (15 + y)/y` ....[From (i)]
⇒ `sqrt(3)y` = 15 + y
⇒ `sqrt(3)y - y` = 15
⇒ `y(sqrt(3) - 1)` = 15
⇒ y = `(15)/(sqrt(3) - 1)`
⇒ y = `(15)/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1`
= `(15(sqrt(3) + 1))/(3 - 1)`
= `(15(sqrt(3) + 1))/(2)"cm"`
⇒ x = `15 + (15(sqrt(3) + 1))/(2)`
= `(30 + 15(sqrt(3) + 1))/(2)`
= `(15(2 + sqrt(3) + 1))/(2)`
= `(15(3 + sqrt(3)))/(2)`
= `(15sqrt(3)(sqrt(3) + 1))/(2)`.
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