Calculate the surface and the angle of the figure below.

[catala]

Homework Statement

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

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

Homework Equations

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

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

Pythagorean Theorem

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

Help please

 

[Mark44]

Homework Statement

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

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

Homework Equations

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

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

Pythagorean Theorem

The Attempt at a Solution

I have calculated from differents methods, one with one rectangle and triangle.

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.

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

Help please

 

[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?

 

[Mark44]

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°.