Practical trigonometry

We have a huge Eucalyptus tree next to our garage. the garage is across the road from the house and because it's so windy today (gusting to 40mph) while watching the big tree sway in the wind I wondered if it would hit the house if it came down. Hmmm.

I remember from taking trig in high school there are multiple ways of accurately calculating the height of a tree, hill or any tall object. Here's one of them:

 

O=AtanθO=Atan⁡θ where O is the opposite length of a right-angled triangle, and A is the adjacent length. Hence O is the height of the tree, and A is the distance from the tree.

All right then....enough of that.

I called a tree guy I know, here's how a guy who cuts down trees for a living does the calculation.

"Take a square piece of paper or cardboard, and fold it along its diagonal to form a 45 degree angle. Start at the base of the tree and pace away from it, turning occasionally to sight the top of the tree. To sight it, sit down on the ground (to minimize error), hold the paper so one flat side is parallel to the ground and you are looking up along the diagonal edge. Repeat this enough times to find the place where you are able to sight the top of the tree along the 45 degree line of your paper. You are now as far away from the tree as it is tall."

He says he can estimate the height within 10% every time.

I got a large triangle out of my old sailing navigation kit and went to work in the wind. Here's what I came up with:

The Eucalyptus is 206 feet tall, give or take ten percent.  If it falls towards the house, it will miss the house by around 20 feet but it will take down all the power lines, the cable line, the fence, the gate and a couple of smaller trees. It will block the road for hours until it's cut up.

The only good news is that it would supply around 20 cords of wood or more for our stove.