# Thread: Complex (for me) Right Triangle

1. ## Complex (for me) Right Triangle

Hello; essentially, I’m new to trig, and cannot figure out the problem below. I am able to manage what I would consider straight forward Right Triangles, but this one has a wee bit on the end, that I do not know how to sort out.
I believe using SOH-CAH-TOA (new term for me) is the right method to solve this, as (again I believe) Sine requires 2-angle/sides, and Cosine requires 1-angle/side…?
Best I was able to manage was figure out Angle D (180-90-30 = 60).
Additionally, I thought that the red forward arrows (hash-marks) meant that two-sides were equal, but I’m not sure in this case. I would appreciate knowing what these marks indicate.
I would appreciate any help on how to approach and solve this problem. I’m one of those people who require a good example to draw upon for future reference.
Thanks in advance for the forums help!

Hello; essentially, I’m new to trig, and cannot figure out the problem below. I am able to manage what I would consider straight forward Right Triangles, but this one has a wee bit on the end, that I do not know how to sort out.
I believe using SOH-CAH-TOA (new term for me) is the right method to solve this, as (again I believe) Sine requires 2-angle/sides, and Cosine requires 1-angle/side…?
Best I was able to manage was figure out Angle D (180-90-30 = 60).
Additionally, I thought that the red forward arrows (hash-marks) meant that two-sides were equal, but I’m not sure in this case. I would appreciate knowing what these marks indicate.
I would appreciate any help on how to approach and solve this problem. I’m one of those people who require a good example to draw upon for future reference.
Thanks in advance for the forums help!

You can use similar triangles to solve this.

tan(30) = opp/adj = 7/(BC). tan(30) is a common value on the unit circle and it's sqrt(3)/3

$BC = \frac{7}{tan(30)} = \frac{7\sqrt(3)}{3}$

$\frac{7\sqrt(3)}{3} = 7/x$ and so $x = \sqrt3$

now you can work out the two hypotenuses with Pythagoras

Keep in mind, I’m still relatively new at this, but let’s see what happens...

Step 1: Find value for BC

7/tan(30) = 7 sqrt(3) = 12.1243

BC = 12

---------------------------------

Step 2: Find value for BD

BD = 12^2 + 7^2 = sqrt(193) = 13.8924

BD = 14

---------------------------------

Now I gather that the red arrows indicate that each side of the smaller right triangle are equal at 1mm each. That being the case, then I’m confused about the red arrows on the larger right angle...as that would then suggest that BC is also equal to BD?

Regardless, I’ll move forward, and hopefully someone can help me understand further.

We now end with:

Step 3: AB + BD = AD

1 + 14 = 15