Inner Angle of an Triangle

Hello, please look at the attached image. H and ang are given and I would like to calculate B

Moreover I need all angles from 0 to 180° at a fix H, lets assume H=6. and ang=0.5 ... 179.9.

How can I program this in matlab ? :-(

Answer 1

[Torsten]

tan(ang/2) = (B/2) / H

thus

B = 2 * H * tan(ang/2)

Does that help ?

Best wishes

Torsten.

Answer 2

[Walter Roberson]

Divide into two right triangles and use trig. "Adjacent", or x, would be H, "opposite", or y, would be B/2, theta would be ang/2. By trig, x/y = tan(theta), so x / tan(theta) = y . Substiting back in, B/2 = H / tan(ang/2) and so B = 2 * H / tan(ang/2)

Now you can vectorize, B = 2 .* H ./ tan(ang./2) . ang can be a vector

Caution: tan() expects radians. You need to convert, or you need to use tand()

[Marv]

Hm ok, how can I put this in a loop and save all values:

 angles from 0 to 180° at a fix H, lets assume H=6. and ang=0.5 ... 179.9.

into a vector ?

Something like this ?

H=6; for(ang=1:1:179) B = [ang H 2 * H / tand(ang/2)]'; end

[Torsten]

ang=1:1:179;
H=6.0;
B=2*H*tand(ang/2);
plot(ang,B);

Best wishes

Torsten.

Answer 3

[Marv]

If I calculate this, I get for H=6 the result:

B = 4,97056274847714

for an angle at 45.

But it should be B=6 ??

[Torsten]

You mean B=12 for ang=90 and H=6 ?

That's what you get by the formula.

Best wishes

Torsten.

[Walter Roberson]

Make sure you are using tand() instead of tan() if you are specifying the angle in degrees.