If its a right triangle, use SoH CaH Toah, or Sine = opposite over hypotenuse, Cosine = adjacent over hypotenuse, and tangent = opposite over the adjacent. So, to solve your problem, if "c" is the right angle, Tan(a) = bc/ca or Tan(30)=5/ca or .577 = 5/ca or (.577)*ca = 5 or ca=5/(.577) or ca= 8.66
tan(23) = 1.58815308
a equals 5
5/9=x/27 135=9x 15=x
1/5+1/11 = 11/55+5/55 = 16/55
If its a right triangle, use SoH CaH Toah, or Sine = opposite over hypotenuse, Cosine = adjacent over hypotenuse, and tangent = opposite over the adjacent. So, to solve your problem, if "c" is the right angle, Tan(a) = bc/ca or Tan(30)=5/ca or .577 = 5/ca or (.577)*ca = 5 or ca=5/(.577) or ca= 8.66
tan(23) = 1.58815308
x equals 4 over 35
a equals 5
x = tan-1(5) = 78.69 degrees
5/6=x/30 150=6x 25=x
5/9=x/27 135=9x 15=x
x/5 = 8/21 x = 8/21 * 5 x = 1.904761905
1/5+1/11 = 11/55+5/55 = 16/55
-2 plus 5 equals +3
12 + 5 equals...
x/5 = 3 Multiply both sides by 5 to find the value of x:- x = 15