# Finding the Height of Two Classical Triangles Shortcut

There are two classical trigonometry problems where you are asked to find the height of an oblique triangle. The problem is quite tough because you have to go through series of calculations.

These calculations require your skill in algebra. That’s the common fall down of student because they thought after they’re already done with their heavy algebra lessons.

Mathematics is a continuous learning process, if you want to be the best, you need to give time to it. According to the experts, 2 hours of Math a day would be fine.

Now, this post is about finding the height of two classical triangles. I want to share it because there is a shortcut formula to easily solve it. Below are the two common triangles.

In figure 1, the height of the triangle can be solved using the following formula,

$h=\displaystyle\frac{a}{cotA+cotB}$

In figure 2, the height of the triangle can be found by extending $\overline{EF}$  and dropping the perpendicular from $G$ to $H$.

The height of the triangle can be found by the following formula.

$h=\displaystyle\frac{a}{cotE-cotF}$

Applications:

Worked Problem 1:

A flagpole is standing at the top of the building with a height of $50 m$. Observer A measures the angle of elevation of the top of the pole and found to be 45°. 200 meters farther, observer $B$ that formed a straight line with the pole and observer $A$ measure the angle of elevation of the top of the pole and found to be 30°. What is the height of the flagpole? (Assume that the observers are both on the level ground.)

Solution:

Again, this is Trigonometry; we need to draw the figure like shown below.

The image resembles what we have in figure 1. Thus we can use the formula to solve for the height.

$50+h=\displaystyle\frac{a}{cotA+cotB}$

Why $50+h?$, because the H in the formula is the height from the ground. So we need to include the height of the building.

$50+h=\displaystyle\frac{200}{cot30+cot45}$

$50+h=\displaystyle\frac{200}{\sqrt{3}+1}$

By rationalizing the denominator we have,

$50+h=\displaystyle\frac{200}{\sqrt{3}+1}\cdot\displaystyle\frac{\sqrt{3}-1}{\sqrt{3}-1}$

$50+h=\displaystyle\frac{200(\sqrt{3}-1)}{(\sqrt{3})^2-1^2}$

$50+h=\displaystyle\frac{200(\sqrt{3}-1)}{3-1}$

$50+h=\displaystyle\frac{200(\sqrt{3}-1)}{2}$

$50+h=100(\sqrt{3}-1)$

$h=100(\sqrt{3}-1)-50$

$(100\sqrt{3}-150) m$    or    $23.21 m$

Sample Problem 2:

An airplane is flying with a constant altitude of 1500 m above the ground. Rian saw the airplane above her with an angle of elevation of 60 degrees, after 15 seconds the angle of elevation of the airplane from Rian is 30 degrees. What is the speed of the plane in kph?

Solution:

Tough right? Again we start by drafting the image.

However, we translate the image in the following manner.

That image now is similar to our figure 2 but flipped, thus we can use the second formula to find the distance traveled by the plane. Remember that the question is asking for the speed of the plane. Not the height. The height is given.

$h=\displaystyle\frac{a}{cotB-cotA}$

By property of transversal, $\angle{CAD}=60$ degrees  and  $\angle{CBA}=30$ degrees

$1500=\displaystyle\frac{a}{cot30-cot60}$

$1500=\displaystyle\frac{a}{\sqrt{3}-\frac{\sqrt{3}}{3}}$

$a=1500\cdot\displaystyle\frac{2\sqrt{3}}{3}$

$a=1000\sqrt{3}$

That is now the time it takes for the plane to travel in 15 seconds.

Solving for the rate of the plane,

$R=\displaystyle\frac{D}{t}$

$R=\displaystyle\frac{a}{t}$

$R=\displaystyle\frac{1000\sqrt{3}m}{15s}\cdot\displaystyle\frac{3600 s}{1 hr}\cdot\displaystyle\frac{1km}{1000m}$

$R=240\sqrt{3} kph$