If A=130° and x=sinA+cosA, then,
x>0
x<0
x=0
x≤0
Explanation for correct option:
Finding the value:
Given, A=130°
Now,
x=sinA+cosAx=sin130°+cos130°x=sin(180°–50°)+cos(180°–50°)x=sin50°–cos50°x=sin50°–cos(90°–40°)x=sin50°–sin40°....(1)
In equation
x=2sin(50°–40°)2cos(50°+40°)2x=2sin5°cos45°sinC–sinD=2sin(C–D)2cos(C+D)2
As we know that sin5°&cos45° are positive,
Therefore, x is also positive & varies from x>0
Hence, correct option is (A)