ABC is a right angles triangle at B. The hypotenuse AC is four times the perpendicular BD drawn to it from the opposite vertex and AD

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. None of these

Answer

A

Explanation

Let BD = P and DE = x
`rArr AC = 4P`

Let E be mid-point of AC.
Then, AE = EC = BE = 2P
In `DeltaBDE, (BE)^(2)=(BD)^(2)+(ED)^(2)`
`rArr (2P)^(2)=(P)^(2)+x^(2)`
`rArr 4P^(2) = P^(2) + x^(2)`
`rArr 3P^(2)=x^(2) rArr x = sqrt(3)P`
Now, `AD = 2P - x = 2P - sqrt(3)P = P(2 - sqrt(3))`
`DC = 2P + x = 2P + sqrt(3)P = P(2 + sqrt(3))`
In `Delta`BAD, tan A = `(BD)/(AD)=(P)/(P(2-sqrt(3)))`
`=(1)/(2-sqrt(3))xx(2+sqrt(3))/(2-sqrt(3))=2+sqrt(3)=tan75^(@)`
`tan alpha=(AD)/(BD)=(P(2-sqrt(3)))/(P)=2-sqrt(3)=2-sqrt(3)=tan 15^(@)`
`rArr alpha = 15^(@)`
As, `Delta`ABC is right angled at B, from figure `alpha + beta = 90^(@)`
`rArr 15 + beta = 90^(@) rArr beta = 75^(@)`
In `Delta ABC, angle A + angle B + angle C = 180^(@)`
`rArr 75^(@) + 90^(@) + angle C = 180^(@)`
`rArr angle C = 180^(@) - 165^(@) = 15^(@)`
`therefore` One of the a cute angle is `15^(@)`