Question

find the value of x in the given figure.. using applications for trigonometry.. ​

Answer 1

Answer:

let BD = y

in ∆ ABD, tan 60 = x/y

=> √3 = x/y

=> x = √3y (1)

in ∆ ABC, tan 30 = x/(40+ y)

1/√3= x/40+y

=> x = 40+y/√3 (2)

LHS of equations 1 and 2 are same, hence we can equate RHS of 1 and 2

=> √3y = 40+y/√3

=> √3×√3y = (40+y)

=> 3y = 40y + y

=> 2y = 40

=> y = 20 m

x = 20√3 [ by substituting value of y in equation 1)

option 4