Lesson video

Hello, everybody, I'm Mr. Gratton and welcome to this trigonometry lesson.

Thank you so much for joining me.

In this lesson, where we will have a look at the origins of the tan ratio from sides of a right-angled triangle.

Pause here to check that you are familiar with some of the really important keywords that we'll be using today.

Let's first look at triangles from the unit circle and see how we can use them along with ratio tables to find the lengths of sides of a triangle.

Izzy recalls that there are several different types of right-angled triangle that can be drawn on a unit circle.

And on some of them, the side that is adjacent to the angle theta degrees will have a length of one unit.

This is different from some of the other triangles that you may have seen in the past where it is the hypotenuse that has length of one unit instead.

On these triangles, with an adjacent length of one unit, the length of the side opposite angle theta lies on a tangent to the unit circle.

Where else in this lesson might we expect to see the word tangent again? For different angles or values of theta, Izzy can record the length of the side opposite to theta, the one on the tangent.

Across all different values of theta that Izzy will measure today, the side adjacent to theta will always remain one unit.

So, if the angle theta is 40 degrees, then the length of the side opposite angle theta is approximately 0.

84 units.

Izzy does this again, but this time with a triangle whose angle theta is 30 degrees.

Then the length of the side opposite it is approximately 0.

57 units.

Note, that all of these values are just approximations as it's pretty difficult to read off of this graph more precisely than that.

And Izzy does this one more time, this time for a triangle whose angle theta is 25 degrees.

Then the length of the side opposite that's 25 degree angle, is approximately 0.

46 units.

As you can see as theta changes, so does the length of the side that is opposite theta.

But what Izzy has just recreated is this, the table of trig values specifically for the tan column.

Wait, why does the word tan sound familiar? Oh, that's right.

It's short for the word tangent.

And the tangent was where the side opposite theta always lay upon on that unit circle.

We can use the tan ratio on this table to find the length of the side opposite that angle.

And so, for this particular triangle, the one with an adjacent side of one unit to the angle of 35 degrees, the length of the side opposite the 35 degree angle is the tangent of 35 degrees, which is 0.

7.

So the length that is the tangent of 35 is 0.

7 units long.

So let's get some quick practise in for using the tangent column in this table.

Pause now to use this table of trig values to find the length of the side labelled K centimetres.

And so, I look at the tangent column, I look at the 15 degree angle which has a length of 0.

268 centimetres.

And pause again to use this table of trig values to find the length of the side labelled W centimetres.

Sticking with the tan column, this time let's have a look at the 65 degree angle.

Therefore, we've got a length W centimetres that is 2.

145 centimetres.

Pause here to consider which of these statements are correct for a triangle with an adjacent side of length one centimetres.

The larger the size of an angle, the longer the length of the side opposite it.

The larger the size of an angle theta degrees, the greater the size of tan theta.

We can also tell that C is true by looking at the table of trig values.

Helpfully, Izzy's comment is correct.

It is also possible to find the length of a side opposite any angle for any right-angled triangle.

Even if the side adjacent to the angle isn't one unit, as long as you know the length of the adjacent side and the angle as well.

To do this, we need to sketch a triangle similar to the one that we want to find the opposite side of.

This similar triangle of course, must have the same angles but we must ensure that it has an adjacent length of one unit.

So let's sketch a triangle similar to this one.

It's going to have an angle of 50 degrees and an adjacent of one centimetre.

Because my adjacent is one centimetre, the side opposite 50 degrees lies on the tangent of a unit circle.

And so its length is the tangent of 50 degrees, which I can find on my table of values to be 1.

192.

I can also see that one centimetre and 10.

2 centimetres are both corresponding sides, both adjacent to the angle 50 degrees.

And so the scale factor multiplier from our sketch to our original triangle is a multiply by 10.

2.

And so the length of the side opposite our 50 degree angle is 10.

2 times by 1.

192 which is 12.

16 centimetres when rounded.

And so, for this next check, a triangle similar to this one will be sketched.

The side of the sketch that is adjacent to the 55 degree angle has a length of one centimetre.

Pause now to use the table of trig values in order to find the length of the side labelled A.

We know the length of the side adjacent to the angle and we want to find the length of the side opposite the angle.

Therefore, we have to use the tangent column and look at the angle 55 degrees, and therefore, A has a length of 1.

428 centimetres.

And so, now that we know the lengths of the two sides of this sketched triangle, pause now to use similar triangles to find the length of the side labelled B.

We know the scale factor is 20 by looking at the corresponding adjacent sides of both of these triangles.

Therefore, 20 times 1.

428 equals 28.

56 centimetres.

We can also do the opposite using the tan ratio in order to find the length of a side adjacent to an angle when we know the length of the opposite angle.

This is the reverse of before when we knew the adjacent side but not the opposite side.

For questions like this, it helps to use a ratio table to visualise how the sides of these similar triangles are related.

So as with before, sketch a similar triangle, the adjacent remains at one centimetre because we always want our similar triangle to be one from the unit circle.

And by ensuring that our sketch has an angle of 35 degrees, we can use the table of trig values to find that the tangent of 35 degrees is 0.

7.

We can then put the lengths of our sketch into the ratio table.

The scale factor multiplier between our sketch and the original triangle is not as straightforward as before.

The scale factor is 16 over 0.

7.

By placing the lengths of the sides of our original triangle into our ratio table, we can also see that the length of the side adjacent to the 35 degree angle, the one currently labelled X, is one times the fraction 16 over 0.

7.

Using our calculator we can see that 16 over 0.

7 is 22.

86 centimetres after rounding.

And for this final set of checks, here is a triangle and its sketch with an adjacent side of length one centimetre.

Pause here to use this ratio table to find the length of the side labelled A.

The tangent of 25 gives 0.

466 centimetres.

As a fraction, what is the scale factor from the triangle with an adjacent of one centimetre to the other triangle? Pause here to use the ratio table to inform your answer.

And the multiplier from 0.

466 to 40 is 40 over 0.

466.

And finally, pause here to use a calculator.

Find the length of the side labelled X, the side adjacent to the 25 degree angle.

And our answer is one times 40 over 0.

466, which is 85.

84 centimetres.

Okay, onto the practise tasks.

For question one, pause here to fill in the blanks on the triangle and in the ratio table by referring to the table of trig values.

By considering corresponding sides between the two triangles, find the length of the side labelled A centimetres.

And for question two, pause here to fill in all the blanks on this triangle and ratio table and find the value of B centimetres by considering corresponding sides.

And pause here again, for this next pair of similar triangles for question two.

And for question three, you'll have to sketch the similar triangles yourself.

Each sketch has to have a side adjacent to the labelled angle of one unit.

Use these sketches to find the length of the missing side of each triangle.

And pause here to use each of your sketches to find the length of the missing side of each of these triangles.

And a very well done for persevering through all of these questions.

And now onto the answers.

Pause here to check and compare all of the values that you filled in with the ones on screen.

And a very well done if you found the length of the side A centimetres to be 4.

958 centimetres.

And for question two, pause here to check that all of your values match with the ones on screen now.

And well done if you found the lengths of B centimetres to be 26.

0965 centimetres, and C centimetres to be 110.

848 centimetres.

And finally question three, the length of the opposite side labelled B centimetres is 128.

7 centimetres.

And C centimetres is 1.

04896 centimetres.

And for the side that is adjacent to the 40 degree angle, we have a length of 22.

05 centimetres.

Let's see if we can spot a relationship between the side opposite an angle and the side adjacent to that exact same angle.

Izzy recalls that when she learnt about the sine and cosine ratios that there was a relationship where the length of one of its sides was the product of H, the hypotenuse, and either the sine or cosine of an angle on that triangle.

So let's have a look on the left hand triangle.

Define the length of the side opposite theta, we start with the hypotenuse, then multiply it by sine of theta, so that the hypotenuse times by sine theta is the length of the opposite.

Similarly, for this right-hand triangle, to find the length of the side adjacent to theta, we start with the hypotenuse, then multiply it by the cosine of theta, and so the hypotenuse times by cosine theta is the length of the adjacent of that triangle.

These are two really helpful relationships involving sine and cosine.

But is there a similar relationship involving tan between the sides adjacent and opposite an angle in the triangle? Well, let's look at this triangle with length adjacent to theta of one centimetre.

This similar triangle can have an adjacent of any length.

For the moment, let's call it K.

The angle theta can be any angle less than 90 degrees, so it is appropriate for a right-angled triangle.

And if all of this is true, then the side opposite to the angle theta will be K times by the tangent of theta.

Let's have a look why.

For this triangle from the unit circle, I know that the length of the side opposite theta will be tangent theta.

The side adjacent to theta on both triangles are related by the scale factor of K.

Therefore, the side opposite theta, but this time on the other similar triangle is going to be K multiplied by the tangent of theta.

And so, rather than having to sketch a similar triangle for every single question, we can instead look at the relationship between two sides, the adjacent and the opposite in only one triangle.

There is a multiplicative relationship between these two sides, the adjacent and the opposite sides on all right-angled triangles.

So, I take the adjacent and multiply it by the tangent of theta.

This always gives the length of the side opposite to theta.

Therefore, the relationship between these two sides can be shown by the opposite side equals the adjacent side times by the tangent of theta.

For this check, pause here to use this relationship in order to find the values that should go into these gaps.

I always start with the adjacent, which in this case is 68 centimetres and multiply it by the tangent of 80 degrees to get the length of the opposite side to the angle 80 degrees.

And hence, pause here to find the length of the side opposite the angle 80 degrees using this table of trig values.

I calculate 68 times by 5.

671 to be 385.

628 centimetres.

So we've looked at finding the length of the side opposite an angle given the side adjacent to it, but it is also possible to do the reverse.

Find the length of the side adjacent to an angle given the length of the side opposite to it.

We can do this by substitution and rearrangement to solve an equation.

For example, this is our relationship and we know that the side opposite to the 30 degree angle is 78 centimetres.

And we know that the angle is 30 degrees itself.

We can use our table of trig values to spot that the tangent of 30 equals 0.

577.

Now that we have an equation, we can rearrange it by dividing through both sides by 0.

577 to get the adjacent equals 78 divided by 0.

577.

Typing this in on my calculator gives approximately 135.

18.

Therefore, the length of the adjacent is 135.

18 centimetres.

For this check, pause here to use the relationship to find the values that should go into these two gaps.

The side opposite the 75 degree angle is 105 centimetres.

So 105 equals the unknown adjacent times by the tangent of 75 degrees.

Pause here again to answer, what goes into these gaps when trying to rearrange and solve this equation.

We are given that the tangent of 75 degrees is 3.

732.

When trying to rearrange and solve this equation, we need to divide through by the tangent of 75 and so we can divide through by 3.

732 on both sides of the equation.

By solving this equation, pause here to find the length of the side adjacent to that 75 degree angle.

The length of the side is 28.

135 centimetres after rounding.

Okay, onto the next set of practise questions.

For question one A, pause now to fill in every gap in the method to help you find the length of the side opposite 40 degrees.

And pause again to find the length of the side opposite the labelled angle in each of these three triangles.

And for question two, it is the adjacent side that is unknown instead of the opposite side.

Pause here to substitute, rearrange, and solve the equations created from the relationship between the adjacent side and the opposite side of each of these triangles.

And pause here again, for parts C and D.

And for D, it might be helpful to sketch your own triangle.

Great work on everything that you've done so far.

Here are the answers to question one.

Pause now to compare to the information on screen, your method and the length of the opposite sides that you found.

And for question two, pause here to compare to the answers on screen, the lengths of the sides adjacent to the marked angles.

Now that we know there is a relationship between the side adjacent to and the side opposite to an angle in a right-angled triangle, can we use a calculator to more efficiently and effectively calculate the lengths of these sides using the tangent function on a calculator? Let's have a look.

Jacob seems pretty confident that he can use this table of trig values to help him find the lengths of missing sides of triangles whose angles are multiples of five.

For this triangle, he takes the side adjacent to the 55 degree angle and multiplies it by the tangent of 55 degrees.

We can use this table of trig values to figure out that the tangent of 55 degrees is 1.

428.

And so, he gets the length of the opposite side to 55 degrees of 82.

824 centimetres.

But this method only works for angles that are included in this table of trig values.

For this triangle, he'll have to use his calculator to find the tangent of 61 degrees as the angle 61 degrees is not in that table of trig values.

And so yet again, he starts with the adjacent side this time of 100 centimetres and multiplies it by the tangent of 61.

In order to find the value of the tangent of 61, he types it into the calculator to get 1.

80, et cetera.

He then types this number into his calculator again in order to multiply it by 100.

Therefore, the length of the side opposite that 61 degree angle is 180.

4 centimetres.

Note that the calculator gives you the sine, cosine and tangent of an angle to far more precision or many more decimal places than the table of trig values.

It is absolutely okay to round your answer at the end of a calculation only though to a suitable number of decimal places or significant figures.

The calculator was meant to make things quicker, right? But as Jacob says, typing out that whole value that represents the tangent of 61 is quite long-winded.

Can we use the calculator in a more efficient way? Well, of course, we can.

Grab your calculator and let's see if we can do this question together but efficiently.

Before we start though, make sure that you can see the letter D at the top of the screen.

And so, we can start with the side adjacent to the angle.

So type in 100, then let's multiply it to the tangent of, and note, the open bracket will appear automatically, to the tangent of the angle 61 degrees.

Note, that you do not need to type in a degrees symbol yourself as we're already in degrees mode.

Don't forget to write down the close bracket yourself.

And then, when you've typed all of this in, press execute to get your answer.

And with only nine buttons pressed in total, not the many more that Jacob had to do when he calculated the side of this triangle.

It's a fairly straightforward calculation to type into your calculator if we want to find the length of an opposite side when given the length of an adjacent one.

But Jacob is correct, I suppose about the opposite calculation.

We do need to perform a few more steps first.

We need to rearrange an equation and start solving it when trying to find the length of an adjacent side if given the opposite side.

But it is still just as possible to efficiently use a calculator to find the length of an adjacent side once our rearrangement steps are complete.

So for this triangle, we start with the adjacent side currently at an unknown length of K.

Then we multiply it by the tangent of the angle 64 degrees which equals the side opposite that angle with a length of 45.

We then start to solve the equation by dividing through by tangent of 64.

We can do this with the expression, the tangent of 64 rather than its decimal equivalent.

This will allow us to type in, 45 over tangent of 64 into our calculator in one big step.

And here's how.

It is always important to start with the fraction button in order to set up the fraction before inputting either the numerator or denominator.

After pressing the fraction button, the calculator defaults to being on the numerator.

So let's type in the length of the opposite side 45.

Then we press the down arrow button in order to transition to the denominator, the number we are dividing by.

And now we can type in the tangent of the opposite angle, which we preserved from our previous calculations.

We want to type in the tangent of the angle 64 and don't forget the close bracket.

We can then press the execute button to get a length of 21.

95 centimetres.

For this check, pause here to match the two triangles to the calculations that find the length of its missing side.

And note, one of these answers is definitely incorrect.

For triangle A, we want to find the length of the opposite side, given that the adjacent has length 200 centimetres.

However, for triangle B, we want to find the length of the adjacent side, given that its opposite has a length of 200 centimetres, instead, it is very rare that we'll see a tangent function as the numerator of a fraction when trying to find the length of an adjacent side or an opposite side.

And for our final check, use a calculator, to find the length of each missing side labelled with an X for both triangles.

And here are the two answers.

And now, onto the final two practise questions.

For question one, pause here to use a calculator and starting with the smallest, put these side lengths of right angle triangles in order of size.

And finally, question two, use a calculator to find the lengths of all missing sides labelled with a letter.

Each triangle is only slightly different from the last.

So see if you can be as efficient as possible with how you use your calculator to answer these questions.

Pause now to do all six.

Great work on using your calculator effectively and efficiently when dealing with all of these tricky calculations.

Pause here to check the correct order for these seven calculations in question one.

And for question two, pause here to check your answers for the lengths of all of these adjacent and opposite sides and compare them with the ones on screen.

And great work, everybody, on getting to grips with the tangent ratio in a lesson where we have explored the idea that the tangent of an angle can help us find the length of a side opposite to that same angle using the tangent ratio for any triangle with an adjacent length of one unit, and also for any triangle of any size by considering the enlargement of a similar triangle of one from the unit circle.

We've also identified the relationship between the side opposite and the angle theta, and the side adjacent to that same angle as the adjacent multiplied by the tangent of that angel equals the length of the side opposite that angle.

And also, we can effectively use a calculator to find the length of either the opposite or adjacent side when the other one is known.

Thank you so much for joining me in today's lesson on trigonometry.

I look forward to seeing you again soon for some more maths, but until that day comes, have an amazing time and goodbye.