A right triangle has perimeter of length 7 and hypotenuse of length `3.` If `theta` is the larger non-right angle in the triangle, then the value of `costhetae q u a ldot` `(sqrt(6)-sqrt(2))/4` (b) `(4+sqrt(2))/6` `(4-sqrt(2))/3` (d) `(4-sqrt(2))/6`

Options

A. `(sqrt6-sqrt2)/4`
B. `(4+sqrt2)/6`
C. `(4-sqrt2)/6`
D. `(4-sqrt2)/6`

Answer

D

Explanation

`"Perimeter "=7=3sintheta+3costheta+3," where " 45^@ltthetalt90^@`
`sintheta+costheta=4/3 ...(i)`
Since `(sintheta+costheta)^2+(sintheta-costheta)^2=2`, we have
`(sintheta-costheta)^2=2-(4/3)^2`

`rArrsintheta-costheta=sqrt2/3 ...(ii)`
From (i) and (ii),`costheta=(4-sqrt2)/6`