Pythagoras Theorem

Find the value of x in the figure below;

Using Pythagoras theorem, we can see that our opposite and adjacent is given but out hypotenuse is given as x. Thus;

$$\begin{gathered} {\mathrm{hypotenuse}}^2={\mathrm{opposite}}^2+{\mathrm{adjacent}}^2 \\ {x}^2=5^2+{12}^2 \\ {x}^2=25+144 \\ {x}^2=169 \end{gathered}$$

Find the square root of both sides

$x=13cm$

If a right angled triangle has equal dimension in two of it sides and the longest side measures 8cm. Find the other sides.

Solution.

From the question our hypotenuse is given as 8cm. However, the opposite and adjacent are not given. Also, we are told opposite = adjacent.

let the adjacent = y; that means opposite = y. Therefore:

Using pythagoras theorem, $$\begin{gathered} {\mathrm{hypotenuse}}^2={\mathrm{opposite}}^2+{\mathrm{adjacent}}^2 \\ {8}^2=y^2+y^2 \\ 64=2y^2 \end{gathered}$$

Divide both sides by 2

$$\begin{gathered} \frac{64}{2}=\frac{2y^2}{2} \\ 32=y^2 \end{gathered}$$

Find the square root of both sides of the equation

$$\begin{gathered} \sqrt{32}=\sqrt{y^2} \\ y=4\sqrt{2}cm \end{gathered}$$

So the opposite is 4√2cm and the adjacent is 4√2cm.

If sinØ = 2/5, Find cosØ and tanØ.

Solution

$SinØ=\frac{opposite}{hypotenuse}=\frac{2}{5}$

This means that the opposite is 2 and the adjacent is 5. Meanwhile, we need to find the adjacent:$$\begin{gathered} {\mathrm{hypotenuse}}^2={\mathrm{opposite}}^2+{\mathrm{adjacent}}^2 \\ {5}^2=2^2+adjacen{t}^2 \\ 9=4+adjacen{t}^2 \end{gathered}$$

Subtract 4 from both sides of the equation.

$$\begin{gathered} 9-4=4+adjacen{t}^2-4 \\ 5=adjacen{t}^2 \end{gathered}$$

Take the square roots.

$adjacent=\sqrt{5}$

Now, we have values for all sides.

$$\begin{gathered} \cos \varnothing =\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} \\ \cos \varnothing =\frac{\sqrt{5}}{5} \\ \mathrm{Tan}\varnothing =\frac{\mathrm{opposite}}{\mathrm{adjacent}} \\ tan\varnothing =\frac{2}{\sqrt{5}} \end{gathered}$$

Rationalize by multiplying the denominator and numerator by √5.

$$\begin{gathered} tan\varnothing =\frac{2\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}} \\ tan\varnothing =\frac{2\sqrt{5}}{5} \end{gathered}$$

Determine if the following is a Pythagorean triple.

1. 7, 12 and 5

2. 8, 15 and 17

Solution

1. To confirm if the series 7, 12 and 5 are Pythagorean triples, take the square of the largest number.

The largest number is 12 and its square is 144.

You should sum the squares of the other two numbers in the series.

the square of 7 is 49

the square of 5 is 25

49+25 = 74

$$\begin{gathered} 144\ne 74 \\ {12}^2\ne 7^2+5^2 \end{gathered}$$

This means that the series 7, 12 and 5 is not a Pythagorean triple.

2. To confirm if the series 8, 15 and 17 are Pythagorean triples, take the square of the largest number.

The largest number is 17 and its square is 289.

You should sum the squares of the other two numbers in the series.

the square of 8 is 64

the square of 15 is 225

64+225 = 289

$$\begin{gathered} 289=289 \\ {17}^2=8^2+{15}^2 \end{gathered}$$

This proves that the set 8, 15 and 17 is a Pythagorean triple.