Archive for the ‘Triangles’ Category

Trigonometry: Right-angled triangles Quesitons

November 22, 2010 Leave a comment

For each set of questions find the value of \mbox{w, x, y and z}

Make sure to write down all the working.  It is a good habit to be rigorous in your answers.

Set 1

Set 2


Set 3

Set 4

Set 5

Categories: Triangles

Trigonometry and Right-angled triangles

November 21, 2010 Leave a comment

Conceptual Understanding

If we find the ratio between a pair of sides on a triangle the value will be a constant even on larger triangles provided the triangles are similar to each other.

The right-angled trigonometry triangle

The hypotenuse is the longest side and is opposite the right-angle

The opposite is the side that is opposite the given angle. (Or the angle we wish to find)

The adjacent is the side next to the given angle.  (It is between the given angle and the right angle)

The ratios SohCahToa


\displaystyle sin \theta = \frac{opp}{hyp}

\displaystyle cos \theta = \frac{adj}{hyp}

\displaystyle tan \theta = \frac{opp}{adj}

When solving a trigonometry right-angled triangle problem we need two pieces of information and then use the relevant ratio to find the third.

There are three possible scenarios.

  1. We are finding an angle
  2. We are finding a side that is a numerator in the ratio
  3. We are finding a side that is a denominator in the ratio


Type 1 (The unknown is the numerator)

We have the hypotenuse and we want the opposite side.  The only ratio with hypotenuse and opposite in it is the sine ratio.

\displaystyle sin\theta = \frac{opp}{hyp}

Let’s fill out the equation

\displaystyle sin 35^o=\frac{x}{17}

\displaystyle \implies 0.374=\frac{x}{17}

\implies x=9.75cm

Type 2 (The unknown is the denominator) 

We have the opposite and we want the adjacent side.  The only ratio with adjacent and opposite in it is the tangent ratio.

\displaystyle tan \theta = \frac{opp}{adj}

Let’s fill out the equation

\displaystyle tan 27^o=\frac{28}{x}

\implies x\times tan 27^o=28

\displaystyle \implies x=\frac{28}{tan 27^o}

\therefore x =54.95mm

Type 3 (The angle is unknown) 

We have the hypotenuse and adjacent side.  The only ratio with adjacent and hypotenuse in it is the cosine ratio.

\displaystyle cos \theta = \frac{adj}{hyp}

Let’s fill out the equation

\displaystyle cos x=\frac{15}{32}

\implies \displaystyle x=cos^{-1}\left(\frac{15}{32}\right)

\implies x =62.0^o


Categories: Triangles

Pythagoras’ Theorem Questions

November 20, 2010 Leave a comment

Set 1: Find the value of a, b, c and d

Set 2: Find the value of a, b, c and d

Set 3: Find the value of a and c in the first diagram and the area of the second diagram

Categories: Triangles

Pythagoras’ Theorem

November 12, 2010 Leave a comment

Definition: The square of the hypotenuse is equal to the sum of the squares of the other two sides.


Right angled triangle with squares

What this means is that if we draw three squares where the length of the sides of the squares are equal to the lengths of the sides of the triangle the area of the big square equals the area of the two smaller squares added together.  (note: The hypotenuse is opposite the right angle and this side produces the largest square)

In the diagram above the largest square is the green square so this implies that c^2=a^2+b^2

It follows that if we are finding a shorter side then we could use a^2=c^2-b^2


Find the length of the hypotenuse.


Here a=3\mbox{ and }b=7

So using c^2=a^2+b^2

We get c^2=3^2+7^2=58

\implies c=\sqrt{58}=7.62cm

Find the length of the missing side 


Since we are finding one of the shorter sides we will use a^2=c^2-b^2

Here b=4\mbox{ and }c=12

We get a^2=12^2-4^2=128

\implies a=\sqrt{128}=11.31cm

Find the value of xThis is a right-angled triangle so let’s use c^2=a^2+b^2 


\implies 50^2=x^2+(2x)^2

\implies 2500=5x^2

\implies x^2=500

\implies x=\sqrt{500}=22.36mm

Does 11cm, 59cm, 60cm produce a right-angled triangle?Use c^2=a^2+b^2 where c=60 since that is the longest length. 

\implies 60^2=11^2+59^2

But 3600\neq 3602, therefore this is not a right-angled triangle.


Categories: Geometry, Triangles