Hello, using SOH CAH TOA (Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj) how could I find Y & X? Would it be SIN (45˚) y/7 or Cos (90˚) x/7?
I tried using the $\sqrt{2}(x)$, 1x, 1x but I received a decimal and am not able to input it: $7/\sqrt{2}$.
How to use SIN, COS, & TAN to solve this equation
algebra-precalculustrianglestrigonometry
Best Answer
Your solution $(\frac{7}{\sqrt{2}}$ or $\frac{7\sqrt2}{2})$ is correct. It's either a problem of the system or it may want the rationalized form $(\frac{7\sqrt2}{2})$.
I would like to introduce another solution to the problem as well.
In a triangle, the angles add up to $180^{\circ}$. So: $\measuredangle{ACB} + \measuredangle{CBA} + \measuredangle{CAB} = 180^{\circ}$. And we know that $\measuredangle{ACB} = \measuredangle{CBA} = 45^{\circ}$, $\measuredangle{CAB} = 90^{\circ}$. In other words, $\triangle{ABC}$ is a right triangle.
Also, notice that $\measuredangle{ACB} = \measuredangle{CBA}$, $\triangle{ABC}$ must be isosceles. Therefore, $AB = AC$ or $x = y$.
Apply Pythagoras theorem in $\triangle{ABC}$: $AB^2 + AC^2 = BC^2 \Rightarrow 2x^2 = 2y^2 = 49 \Rightarrow x = y = \frac{7\sqrt{2}}{2}$