# Calculate the surface and the angle of the figure below.

1. Nov 17, 2012

### catala

1. The problem statement, all variables and given/known data

http://ima.cs-gamers.com.ar//images/490matjhs.jpg [Broken]

I have to calculate the surface from the object and the angle alpha.

2. Relevant equations

$A_{triangle} = \frac{b * h}{2}$

$\frac{a}{\sin \alpha_1} = \frac{b}{\sin \alpha_2}$

Pythagorean Theorem

3. The attempt at a solution

I have calculated from differents methods, one with one rectangle and triangle. And the other form with dividing the object with two triangles.

Dividing the object in two triangles thus leaving:

The triangle with base (640) Can be that the angles of this triangle are 90 , 45 and 45?

Because if they are well the height gives a value of I 640, which in the drawing does not correspond

Last edited by a moderator: May 6, 2017
2. Nov 17, 2012

### Staff: Mentor

This is the simpler way, with a right triangle on the left and a rectangle on the right. It's very easy to get one of the legs of the right triangle. A little right triangle trig will then get you the side opposite the angle α.

Last edited by a moderator: May 6, 2017
3. Nov 18, 2012

### catala

I have done the following:

$\frac{500}{\sin 90º} = \frac{400}{\sin \beta} \to \beta = 53, 13º$

$90º + 53,13º + \alpha = 180º \to \alpha = 36,87º$

$\sin 36,87º = \frac{h}{500} \to h = 300$

$A_{triangle} = \frac{b*h}{2} = \frac{400 \cdot 300}{2} = 60000m^2$

$A_{rectangle} = a * h = 640 * 300 = 192000 m^2$

$A_T = 252000m^2$

Is that correct?

Last edited: Nov 18, 2012
4. Nov 18, 2012

### Staff: Mentor

Looks good.

You could have saved yourself some work by noting that the triangle is a right triangle. You know the base (400 m.) and the hypotenuse (500 m.), so cos(α) = 400/500 = 4/5 $\Rightarrow$ α = cos-1(4/5) ≈ 36.87°.