Answer:
18.975
Step-by-step explanation:
I hope this is right but i did 60 times 4 =240. then I divided 4554 by 240 but i hope this works and can yhu mark me brainliest
☆ given that ,
ABCD is a parallelogram
☆ we know that ,
opposite angles of a parallelogram are equal.
Therefore ,
therefore , ∠B = 3x = 3 ( 26° ) = 78°
thus option ( A ) is correct ~
hope helpful~
When 
![t=\cos A+\sin A\in[-\sqrt{2};\sqrt{2}]\approx[-1.4142;1.4142]](https://tex.z-dn.net/?f=%20t%3D%5Ccos%20A%2B%5Csin%20A%5Cin%5B-%5Csqrt%7B2%7D%3B%5Csqrt%7B2%7D%5D%5Capprox%5B-1.4142%3B1.4142%5D%20)
and there is only one answer t = 1.
For 
both values are correct.
Sin(13π/8) , is in quadrant IV so the angle will be negative, find the reference angle.
<span>sin(13π/8) = -sin(16π/8 - 13π/8) = -sin(3π/8) = -cos(4π/8 - 3π/8) = - cos(π/8). </span>
<span>Use half angle formula for cos(x): </span>
<span>cos²(x/2) = (cos(x) + 1) / 2 </span>
<span>Let x = π/4 </span>
<span>cos²(π/8) = (cos(π/4) + 1) / 2 </span>
<span>cos²(π/8) = (√(2) / 2 + 1) / 2 </span>
<span>cos(π/8) = √(√(2) / 4 + 1/2) </span>
<span>-cos(π/8) = -√((√(2) + 2) / 4)</span>
7 and 1/3 so 7 1/3
Hope this helps
