# Determine length from sketch

I have a simple problem that I need to solve. Given a height (in blue), and an angle (eg: 60-degrees), I need to determine the length of the line in red, based on where the green line ends. The green line comes from the top of the blue line and is always 90-degrees.

The height of the blue line is variable. The angle of the blue line is variable.

Also, I do not know the length of the green dashed-line. Is there a way to figure out the length of the red line without knowing the length of the green?

Any help would be much appreciated!

• $\cos 60=\frac{10}{\color{red}{red}}$
– Mann
May 14 '15 at 19:20

Since you're dealing with right triangle, you can just use cosine function:

$$\cos(\theta)=\frac{blue}{red}$$

Or, substituting $$60^o$$ for $$\theta$$ and 10 for blue:

$$\cos(60^o)=\frac{10}{red}$$

Thus, solving for the length of the red side:

$$red=\frac{blue}{\cos(\theta)}$$

Or

$$red=\frac{10}{\cos(60^o)} = 20$$

Use Law of sines, $$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$$

Let,

$\angle{A}=60^o$

$\angle{B}=90^o$

$\angle{C}=30^o$

$a=?$

$b=?$

$c=10cm$

From Law of sines we have,

$$\frac{a}{\sin{A}}=\frac{c}{\sin{C}}$$

Put the values and find $a$.

Now, from Law of sines we have,

$$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}$$

Put the known values and find $b$.

Congratulations, you are done!

• The problem is that I do not know the length of the green dashed-line. Also, I will need to turn this in to software. Is there a way to help express this in a format I can use (C++/Objective-C, etc)? Much thanks
• Yes, of course! There is a way which is clear in my answer. If you want to know the length of red line without determining the length of green dashed-line then use only, $$\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$$. where, $b$ is the length of red line (to be determined) and $c$ is the length of blue line ($10cm$). $\angle{B}=90^o$ and $\angle{C}=30^o$. Did you see, you didn't find any need to find the length of green dashed line for finding the length of red line. So, therefore, my answer is still an answer. :) May 16 '15 at 10:26