Question

Find the value of x in a right triangle

Answer 1

We have three different ways to the solution.

1.
It's a right triangle and isosceles triangle.
The formula of the hypotenuse of this triangle:


`h=x\sqrt2`

therefore


`x\sqrt2=210\ \ \ |\cdot\sqrt2\\\\2x=210\sqrt2\ \ \ |:2\\\\\boxed{x=105\sqrt2}`

2.
Use the Pythagorean theorem:


`x^2+x^2=210^2\\\\2x^2=44,100\ \ \ \ |:2\\\\x^2=22,050\to x=√(22,050)\\\\x=√(11,025\cdot2)\\\\x=√(11,025)\cdot\sqrt2\\\\\boxed{x=105\sqrt2}`

3.
Use trigonometric function:


`\sin45^o=(x)/(210)\\\\\sin45^o=(\sqrt2)/(2)\\\\(x)/(210)=(\sqrt2)/(2)\ \ \ |cross\ multiply\\\\2x=210\sqrt2\ \ \ \ |:2\\\\\boxed{x=105\sqrt2}`

Answer: b. 105√2