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12Feb

How to “solve” a right-angle triangle

12 Feb, 2024 | Articles | 0 Comments | Return|

Painting depicting Pythagoras by J.A. Knapp
Author: Amir Nasabzadeh

This article is intended as a comprehensive guide on how to “solve” a right-angle triangle. I wrote this article because in our field of solar installation we use Pythagoras Theorem a lot to determine the pitch of roofs and the spaces between panels on flat roofs. In future articles I will explain about this further. I hope you find it helpful.
Please let me know if you have questions or comments in the comments section.

The concept of “solving” a right-angle triangle involves calculating all angle measurements and side lengths.
A triangle has six pieces of information. Three angles and three sides.
Having two of these pieces of information, and using Pythagoras Theorem and trigonometric functions allows for the determination of the missing four.

Example 1

We have a right-angle triangle with given:

Angle:

∡ A = 20^{\circ}

$∡ A = 20^{\circ}$
Side:

a = 8

$a = 8$

First, it is very easy to calculate the missing angle “angle B”. As we know when angle A is added to angle B the answer is 90°.

Therefore:

\begin{array}{l} ∡ A = 20^{\circ} \\ ∡ B = 90 - ∡ A \\ ∡ B = 90 - 20 = 70 \\ \underline{\underline{∡ B = 70^{\circ}}} \end{array}

$\begin{array}{l} ∡ A = 20^{\circ} \\ ∡ B = 90 - ∡ A \\ ∡ B = 90 - 20 = 70 \\ \underline{\underline{∡ B = 70^{\circ}}} \end{array}$

This part of equation was very easy.

Now let’s say we would like to find out the length of side across from angle $∡ A$ $∡ A$ which is side ‘a’ and the length of the side of across from angle $∡ C$ $∡ C$ which is side ‘c’.
We can now pick one of the three commonly used terms in trigonometry – Sine, Cosine and Tangent.
I am not going to go into the full details on this topic as is a huge topic by itself, but they can help us to study the relationships between the different lengths and angles in a triangle.

Pythagoras Theorem Sides (SOH CAH TOA)

SOH CAH TOA:
SOH = Sine is Opposite over Hypotenuse $= \sin θ = \frac{Opposit}{Hypotenuse}$ $= \sin θ = \frac{Opposit}{Hypotenuse}$

CAH = Cosine is Adjacent over Hypotenuse $= \cos θ = \frac{Ajacent}{Hypotenuse}$ $= \cos θ = \frac{Ajacent}{Hypotenuse}$

TOA = Tangent is Opposite over Adjacent $= \tan θ = \frac{Opposite}{Ajacent}$ $= \tan θ = \frac{Opposite}{Ajacent}$

The inverse of sin = sin^-1, cos = cos^-1 and tan = tan^-1.

It is very important that we label triangle correctly.

Opposite is the side opposite the angle we are looking through:

Adjacent is the side adjacent to the angle we are looking through:

Hypotenuse is the longest side of a right-angle triangle and the Hypothenuse is always opposite the right angle.

As I have explained, in this situation we only have two pieces of information. We know that angle A's opposite side is 8 and the side across from the right-angle C is the hypotenuse (c). Which of the above rules of Pythagoras Theorem Sides should we use to find the is 20°, the unknown length of the hypotenuse?

Yes, we have:
$\begin{array}{l} \sin θ = \frac{Opposite}{Hypotenuse} \\ \sin 20^{\circ} = \frac{8}{Hypotenuse} \\ \sin 20^{\circ} = \frac{8}{c} \end{array}$ $\begin{array}{l} \sin θ = \frac{Opposite}{Hypotenuse} \\ \sin 20^{\circ} = \frac{8}{Hypotenuse} \\ \sin 20^{\circ} = \frac{8}{c} \end{array}$

Then what we should do is we should think of this as being divided by 1: $\frac{\sin 20^{\circ}}{1} = \frac{8}{c}$ $\frac{\sin 20^{\circ}}{1} = \frac{8}{c}$

Then we should cross multiply to get rid of the fraction: $\frac{\sin 20^{\circ}}{1} = \frac{8}{c} = c \sin 20 = 8$ $\frac{\sin 20^{\circ}}{1} = \frac{8}{c} = c \sin 20 = 8$

Then we should divide both sides of the equation by sin^20°: $\frac{c \sin 20^{\circ}}{\sin 20^{\circ}} = \frac{8}{\sin 20^{\circ}}$ $\frac{c \sin 20^{\circ}}{\sin 20^{\circ}} = \frac{8}{\sin 20^{\circ}}$

Cancel out: $\frac{c \sin 20^{\circ}}{\sin 20^{\circ}} = \frac{8}{\sin 20^{\circ}}$ $\frac{c \sin 20^{\circ}}{\sin 20^{\circ}} = \frac{8}{\sin 20^{\circ}}$

we have: $c = \frac{8}{\sin 20^{\circ}} = 23.3904352$ $c = \frac{8}{\sin 20^{\circ}} = 23.3904352$
Rounded up: $\underline{\underline{c = 23.39}}$ $\underline{\underline{c = 23.39}}$
So that comes out to 23.3904352, which is rounded to 23.39.

To find out the final side, side ’b’, (I prefer to go back to the original number we were given) in the case of Example 1 this was for Angle: $∡ 20^{\circ}$ $∡ 20^{\circ}$ and for Side: $a = 8$ $a = 8$ , this is because we rounded up side ‘c’.
So, we have got opposite and adjacent. Again, we should use the above rules of Pythagoras Theorem Sides to find the unknown length. We have TOA ratio function. $\tan θ = \frac{Opposite}{Adjacent}$ $\tan θ = \frac{Opposite}{Adjacent}$
We have: $ \tan 20^{\circ} = \frac{8}{Adacent(b)} o r \tan 20^{\circ} = \frac{8}{b}$ $ \tan 20^{\circ} = \frac{8}{Adacent(b)} o r \tan 20^{\circ} = \frac{8}{b}$
Again divided by 1: $ \frac{\tan 20^{\circ}}{1} = \frac{8}{b}$ $ \frac{\tan 20^{\circ}}{1} = \frac{8}{b}$
And cross multiplying: $ \frac{\tan 20^{\circ}}{1} = \frac{8}{b} ∴ b = \frac{8}{\tan 20^{\circ}} = 21.97981936$ $ \frac{\tan 20^{\circ}}{1} = \frac{8}{b} ∴ b = \frac{8}{\tan 20^{\circ}} = 21.97981936$
Rounded up: $\underline{\underline{b = 21.98}}$ $\underline{\underline{b = 21.98}}$

And now we have solved all the sides and all the angles of our triangle in Example 1.

Example 2:

This example presents a more challenging problem where the side lengths are b=6, c=8, but none of the angle measurements are provided.
So, how should we approach this problem? We have a right-angle triangle with given:

Side:

a = 6

$a = 6$
Side:

c = 8

$c = 8$

Let’s solve the missing side first.
To find the measurement of the third side of a right-angle triangle we use the famous Pythagoras Theorem again using the method below:

c^{2} = a^{2} + b^{2}

$c^{2} = a^{2} + b^{2}$
So, we have:
Side:

a = 6

$a = 6$
Side:

c = 8

$c = 8$

Let’s just put the known side into the equation:
$\begin{array}{l} c^{2} = a^{2} + b^{2} \\ 8^{2} = 5^{2} + b^{2} \\ 64 = 36 + b^{2} \end{array}$ $\begin{array}{l} c^{2} = a^{2} + b^{2} \\ 8^{2} = 5^{2} + b^{2} \\ 64 = 36 + b^{2} \end{array}$

Transposing the equation:
$\begin{array}{l} 64 = 36 + b^{2} \\ 64 - 36 = b^{2} + 36 - 36 \\ 64 - 36 = b^{2} + 36 - 36 \\ 28 = b^{2} \end{array}$ $\begin{array}{l} 64 = 36 + b^{2} \\ 64 - 36 = b^{2} + 36 - 36 \\ 64 - 36 = b^{2} + 36 - 36 \\ 28 = b^{2} \end{array}$

Rearrange:
$\begin{array}{l} 28 = b^{2} \\ b^{2} = 28 \end{array}$ $\begin{array}{l} 28 = b^{2} \\ b^{2} = 28 \end{array}$

The square root of both sids is: $\sqrt{b^{2}} = \sqrt{28}$ $\sqrt{b^{2}} = \sqrt{28}$
Cancel out: $\sqrt{b^{2}} = \sqrt{28} ∴ b = \sqrt{28} = 2 \sqrt{7} = 5.29150262$ $\sqrt{b^{2}} = \sqrt{28} ∴ b = \sqrt{28} = 2 \sqrt{7} = 5.29150262$
and rounded up: $\underline{\underline{b = 5.29}}$ $\underline{\underline{b = 5.29}}$

Now we only need to find the two angles, when we find out one of the angles, we can subtract the angle from 90° to solve the other angle.

Let’s fist solve angle $∡ A$ $∡ A$ .

As I mentioned before, the original values are more accurate.
If our calculation on side b had any mistakes, we would then carry the mistake forward. That is why I like to go back to the original problem. Also, if this side is rounded up, the final calculation will not be accurate.

So, if we position ourself over angle $∡ A$ $∡ A$ , we have got the Opposite side and we have got the Hypotenuse.
The trigonometric function for this is SOH.
$\begin{array}{l} \sin θ = \frac{Opposite}{Hypotenuse} \\ \sin A = \frac{Opposite}{Hypotenuse} \end{array}$ $\begin{array}{l} \sin θ = \frac{Opposite}{Hypotenuse} \\ \sin A = \frac{Opposite}{Hypotenuse} \end{array}$

we have: $\sin A = \frac{6}{8}$ $\sin A = \frac{6}{8}$

Now we want to find out the missing angle, this is when the inverse trig functions come into play in this case the sine inverse because the sin is $(\sin^{- 1})$ $(\sin^{- 1})$ those undo one another:
$\begin{array}{l} \sin A = \frac{ 6}{8} \\ \sin^{- 1} (\sin) A = \sin^{- 1} (\frac{ 6}{8}) \\ \sin^{- 1} (\sin) A = \sin^{- 1} (\frac{ 6}{8}) canceled out \\ A= \sin^{- 1} (\frac{6}{8}) ∴ A = 48.5903778 \end{array}$ $\begin{array}{l} \sin A = \frac{ 6}{8} \\ \sin^{- 1} (\sin) A = \sin^{- 1} (\frac{ 6}{8}) \\ \sin^{- 1} (\sin) A = \sin^{- 1} (\frac{ 6}{8}) canceled out \\ A= \sin^{- 1} (\frac{6}{8}) ∴ A = 48.5903778 \end{array}$
So, rounded up we have: $\underline{\underline{∡ A = {48.59}^{\circ}}}$ $\underline{\underline{∡ A = {48.59}^{\circ}}}$

Now we want to find out angle $\begin{array}{l} ∡ B \end{array}$ $\begin{array}{l} ∡ B \end{array}$ , these two angles add up to 90°.
$\begin{array}{l} ∡ A = {48.59}^{\circ} \\ ∡ B = 90 - 48.59 = 41.41 \\ \underline{\underline{∡ B = {41.41}^{\circ}}} \end{array}$ $\begin{array}{l} ∡ A = {48.59}^{\circ} \\ ∡ B = 90 - 48.59 = 41.41 \\ \underline{\underline{∡ B = {41.41}^{\circ}}} \end{array}$

Now we have found all the angles and all the sides of our right-angle triangle in example 2.

Questions

What does it mean to solve a right-angle triangle?
How can the missing angles and sides of a right-angle triangle be calculated?
What are the common trigonometric functions used to solve right-angle triangles?
How does the Pythagorean Theorem apply to solving right-angle triangles?

Blog APN Elec

How to “solve” a right-angle triangle

Example 1

Pythagoras Theorem Sides (SOH CAH TOA)

Example 2:

Questions

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How to “solve” a right-angle triangle

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