Question

A triangle has one 90 degrees angle and from there one leg is 10.25 long and the other leg is 7.75 long, what are the degrees for the other two angles?

Answer 1

You can start by using Pythagoras to find the length of the third side to give yourself more options (long way) and then finish off by using SOHCAHTOA* to find the angles:

The triangle you describe is something like this:

It is a right angled triangle and we only have one side to find (as I will explain the long way), therefore we can easily use Pythagoras to calculate the length of the missing side:

Pythagoras Theorem states:

`a^2 = b^2 + c^2` where `a` is the length of the Hypotenuse and `b` and `c` are the other lengths of the other two sides.

This can be rearranged to find any side in any right angled triangle, as long as there is only one side missing.

Before using Pythagoras, it can be a great help to label the sides `a`, `b` and `c`:

We are missing side `a`, so we can find this using our equation:

`a^2 = b^2 + c^2`

`a^2 = 10.25^2 + 7.75^2`

`a^2 = 105.0625 + 60.0625`

`a^2 = 165.125`

Remember you are finding `a`, not `a^2` - Square root!

`a = sqrt(165.125)`

`a = 12.850` (rounded to 3DP)

We can now add this to our triangle:

Pythagoras - in this question - is unnecessary but can still be done.

Now we have all of the sides, we can use SOHCAHTOA:

`Sinx = ("Opposite")/("Hypotenuse")`

`Cosx = ("Adjacent")/("Hypotenuse")`

`Tanx = ("Opposite")/("Adjacent")`

The Opposite and Adjacent values depend on which angle you are using:

Opposite is the side across from the angle.

Adjacent is the side on the other side of the angle.

Hypotenuse never changes and is the longest side of the right angled triangle.

Let's label our angles using `A` and `B`:

Let's look at angle `A`:

We could choose any side to use, but let's pretend that we didn't have 12.85 (side `a`) as it saves time in exam conditions to leave the Pythagoras out, which leaves us with side `c` and side `b`.

As we look at angle `A`, we can see that the length on the other side of it is 7.75 (side `c`) and the length across from it is 10.25 (side `b`), meaning we have the values of Opposite and Adjacent.

We can use `Tanx` as:

`TanA = ("Opposite")/("Adjacent")`

Therefore:

`TanA = 10.25/7.75`

`TanA = 1.323` (rounded to 3DP)

At this point you use the inverse of Tan (`Tan^(-1)A`), which can usually be done by pressing the Shift or Alt button on your calculator and pressing the `Tan` button:

`A = tan^(-1)(1.323)`

`A = 52.916°` (rounded to 3DP)

The other angle can either be found by going through the same process (using `Sin`, `Cos` or `Tan`), or by manipulating the fact that the sum of all angles in a triangle must equal 180°:

We have 90° and 52.916°:

`90° + 52.916° = 142.916°`

So we can use the following to find the final angle:

`B = 180° - 142.916°`

`B = 37.084°`