
![\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0 \\\\\\ sin(x)~[2cos(x)-\sqrt{2}]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2} \\\\\\ \measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}](https://tex.z-dn.net/?f=%5Cbf%202sin%28x%29cos%28x%29%3Dsin%28x%29%5Csqrt%7B2%7D%5Cimplies%202sin%28x%29cos%28x%29-sin%28x%29%5Csqrt%7B2%7D%3D0%0A%5C%5C%5C%5C%5C%5C%0Asin%28x%29~%5B2cos%28x%29-%5Csqrt%7B2%7D%5D%3D0%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Asin%28x%29%3D0%5Cimplies%20%5Cmeasuredangle%20x%3D0~~%2C~~%5Cpi%20%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A2cos%28x%29-%5Csqrt%7B2%7D%3D0%5Cimplies%202cos%28x%29%3D%5Csqrt%7B2%7D%5Cimplies%20cos%28x%29%3D%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cmeasuredangle%20x%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D~~%2C~~%5Cfrac%7B7%5Cpi%20%7D%7B4%7D)
now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).
No you cannot use the Law of Sines to solve the triangle.
To use the Law of Sines you need to have at least ONE angle measure.
a/Sin A = b/Sin B
a/Sin A = c/Sin C
b/Sin b = c/Sin C
60 and 2x are vertical angle which means that they are congruent.
2x = 60
Now use the equation to solve for x.
2x = 60
x = 60/2
x = 30
Hope this helps :)
Answer:
11) x = 45
12) x = 30
Step-by-step explanation:
11)
3x + x = 180
4x = 180
x = 45
12)
4x = 5x - 30
4x - 5x = -30
-x = -30
x = 30