The base of an isosceles triangle measures 4 units base angle is equal to `45^(@)`. A straight line cuts the extension of the base at a point M at the angle `theta` and bisects the lateral side of the triangle which is nearest to M.
The length of portion of straight line inside the triangle may lie in the range

Options

A. (2,4)
B. `((3)/(2), sqrt(3))`
C. `(sqrt(2), 2)`
D. `(sqrt(2), sqrt(3))`

Answer

C

Explanation

For the given specification of the triangle, consider triangle ABC as shown in the figure.

Equation of line PM is
`y-1 = "tan " theta(x-1)`
For point Q, solving MP and AC,
`4-x-1 = "tan "theta(x-1)`
`rArr Q-=((3+"tan"theta)/(1+"tan" theta),(1+3 "tan" theta)/(1+"tan" theta))`
`"Area of "DeltaAPQ = (1)/(2)||{:(1,1,1),(2,2,1),((3+"tan"theta)/(1+"tan"theta), (1+3"tan"theta)/(1+"tan" theta), 1):}|| = (1-"tan"theta)/(1+"tan"theta)" " ("As " theta lt pi//4)`
Area of quadilateral BPQC,
`Delta = (1)/(2) xx 4 xx 2-((1-"tan"theta)/(1+"tan" theta))`
`"or "Delta = ((3+5"tan"theta)/(1+"tan" theta))`
`"or "Delta = 5-(2)/(1+"tan" theta)" where " theta in (0, (pi)/(4))`
`therefore Delta in (3,4)`
`(PQ)^(2) = ((3+"tan" theta)/(1+"tan" theta)-1)^(2) + ((1+3"tan" theta)/(1+"tan" theta)-1)^(2)`
`=(4(1+"tan"^(2) theta))/(1+"tan"^(2)theta + 2 "tan" theta)`
`=(4)/(1+(2 "tan" theta)/(1+"tan"^(2)theta))`
`=(4)/(1+"sin" 2theta)`
`"So, (PQ)^(2) in (2,4)`
`therefore PQ in (sqrt(2), 2)`