# Math Help - Finding angles in a right angled triangle where one angle is 1/x or x^2

1. ## Finding angles in a right angled triangle where one angle is 1/x or x^2

Working with a right angled triangle ABC with C as the right angle;
If A = x and B = x^2
What are the values of A and B?
Also,
If A = x and B = 1/x
What are the values of A and B?
Obviously
A + B + 90 = 180
So A + B = 90
So
x + x^2 = 90
and
x + 1/x = 90
The book I am working with has the answers in the back, so I know the answers, but I don't understand how the answers were arrived at.

I can work out x + 2x = 90
3x = 90
x = 90/3
x = 30
2x = 60
But I don't know the way to do the same with the reciprocal and the square:
x + 1/x = 90
1/x = 90 - x which cannot be right.

2. Originally Posted by PhilC
Working with a right angled triangle ABC with C as the right angle;
If A = x and B = x^2
What are the values of A and B?
Also,
If A = x and B = 1/x
What are the values of A and B?
Obviously
A + B + 90 = 180
So A + B = 90
So
x + x^2 = 90
and
x + 1/x = 90
The book I am working with has the answers in the back, so I know the answers, but I don't understand how the answers were arrived at.

I can work out x + 2x = 90
3x = 90
x = 90/3
x = 30
2x = 60
But I don't know the way to do the same with the reciprocal and the square:
x + 1/x = 90
1/x = 90 - x which cannot be right.
$x^2+x-90=0$ is a quadratic, use what ever method you have been taught to solve quadratics.

$x+\frac{1}{x}=90$ multiply through by $x$ to get $x^2-90x+1=0$ which is a quadratic ...

CB

3. Thanks! Now I just need to remind myself of how to solve quadratics.

I'm just trying to relearn all my maths from school, and each time I hit a block it sends me down a completely different line of research. It's great this maths stuff, isn't it!