Video: Pack 5 โ€ข Paper 2 โ€ข Question 21

Pack 5 โ€ข Paper 2 โ€ข Question 21

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Video Transcript

๐ท๐ธ๐น is a triangle. ๐บ is a point on ๐ท๐น. Work out the area of triangle ๐ท๐ธ๐น. Give your answer to three significant figures.

So for this question, we can see that we have a big triangle ๐ท๐ธ๐น, but we also have another line ๐ธ๐บ. And this line cuts our big triangle into two smaller triangles, ๐ท๐ธ๐บ and ๐ธ๐น๐บ. Now the question asks us to find the area of the big triangle ๐ท๐ธ๐น. But unfortunately, it doesnโ€™t look like we have a lot of information about this triangle. We have an angle of 70 degrees and an adjacent side length of 3.4 centimeters. Although it may look like we have two of the side lengths, when we look closely, this measurement here of 2.5 centimeters corresponds to ๐บ๐น, not ๐ท๐น. So we donโ€™t actually have two of the sides of our big triangle.

Now the question asks us to find the area of ๐ท๐ธ๐น. In this situation, the formula that we would usually turn to is the area of a triangle is half times the base times the perpendicular height. Observing the diagram for triangle ๐ท๐ธ๐น, it doesnโ€™t look like we have much luck with this formula, since none of the angles appears to be a right angle, nor does it look like weโ€™ll have an easy time finding the perpendicular height given the information on the diagram.

I need to turn to another formula for the area of a triangle, half ๐‘Ž๐‘ sin ๐ถ, where ๐ถ is one of the angles in the triangle and ๐‘Ž and ๐‘ are the two adjacent side lengths of this angle. Now, weโ€™re getting a little ahead of ourselves here. But itโ€™s good to know what weโ€™re aiming for.

This area formula will be the final stage in a multistep process. Again, we donโ€™t have enough information about triangle ๐ท๐ธ๐น to implement this formula yet. And so all of the preceding steps were to help us gather this information using what weโ€™ve been given on the diagram.

Okay, first, a quick recap of some of the tools that will be helping us along the way. The sine rule can be used in triangles where we have two angles and one of the angles opposite these side lengths. Here, the sine rule can be used to find the angle opposite the other side length. It can also be used in triangles where we have two angles and one of the side lengths opposite these angles. Here, the sine rule can be used to find the side length opposite the other angle.

The cosine rule can be used when we have an angle trapped between two side lengths. In this situation, the cosine rule can be used to find the side length opposite the angle we have. The cosine rule can also be used in triangles where all of the side lengths have been given. In this situation, the cosine rule can be used to find any of the three angles in the triangle.

Given these tools, now letโ€™s walk through the method weโ€™ll be using to answer this question. Step one, to begin, weโ€™ll be using the cosine rule on the smaller triangle ๐บ๐ธ๐น. Hopefully, here you can see we have a situation where we have a side, an angle, and a side given. We can therefore use the cosine rule to find the missing side length, which on our diagram is ๐ธ๐บ.

For step two, weโ€™ll continue on the same smaller triangle ๐ธ๐น๐บ. Using the newly discovered side length, we now have a situation where we can use the sine rule and the side-side-angle case. This will allow us to find the missing angle marked here, which is angle ๐น๐ธ๐บ.

For step three, weโ€™ll be working on the big triangle ๐ท๐ธ๐น. One of the angles for the big triangle is given in the question. Here, weโ€™ll be in a position to find one of the other angles by adding the newly discovered angle ๐น๐ธ๐บ to the other angle given by the question ๐ท๐ธ๐บ. By using the fact that angles in a triangle add up to 180 degrees, weโ€™ll be able to find the third and final angle for our big triangle.

For step four, weโ€™ll be using the sine rule again for our big triangle ๐ท๐ธ๐น. This time, we have the other case of the sine rule, where we have an angle, an angle, and a side. Given the newly discovered information, here weโ€™ll be able to use the sine rule to find the side length opposite the other angle, which is side length ๐ท๐ธ. And here, finally, weโ€™ll be able to find the area of our big triangle ๐ท๐ธ๐น by using the formula half ๐‘Ž๐‘ sin ๐ถ that we mentioned earlier. This formula uses the same criteria as our first application of the cosine rule, which is a known angle trapped between two known side lengths. At this stage in our calculations, we will indeed have this information about our big triangle ๐ท๐ธ๐น. And we will therefore be able to use the formula to find the area of our triangle.

Great! Now that we understand our method, letโ€™s get to the calculations. Here we have the triangle ๐บ๐ธ๐น with the information that we currently know. And here we have the cosine rule in the form thatโ€™s most useful for the side-angle-side configuration. The notation used here is that a lowercase letter represents one of the side lengths for our triangle, and the corresponding uppercase letter represents the angle opposite that side length.

Here, we label our triangle to give some context. We have the side lengths ๐‘Ž, ๐‘, and ๐‘ and the angles which are uppercase ๐ด, ๐ต, and ๐ถ. We have chosen to label in this way because we know all of the information now on the right-hand side of this equation: the side lengths lowercase ๐‘ and ๐‘ and the angle uppercase ๐ด. This will allow us to substitute in the known values and solve to find the unknown side length ๐‘Ž, which is our desired side length on the triangle ๐ธ๐บ.

To make things extra clear, for this first calculation, here weโ€™ve the substitutions weโ€™ll be making. Once we perform the substitutions, our cosine rule looks like this. For this type of question, make sure your calculator or device is set to degrees instead of radians to avoid mistakes.

Now here for our next step of working, youโ€™ll notice that the value of cos 70 isnโ€™t a nice round number. So somewhere later in our calculations, weโ€™ll be approximating. Moving to the next step, we find that the value for ๐ธ๐บ squared is equal to 11.9956 dot dot dot. To find the value for ๐ธ๐บ, we take the square root of both sides of this equation. And of course we ignore any negative solutions to the square root since ๐ธ๐บ is a length.

Now here itโ€™s worth noting, for questions like this, youโ€™re usually given a little bit of room for accuracy on your answer. But try to keep as many decimal places as you can on your calculator, especially early on in your calculations, so you donโ€™t lose too much to rounding.

And so performing a square root on our long decimal number, we find the value for ๐ธ๐บ is equal to 3.4634 dot dot dot centimeters. Great! Weโ€™ve now completed step one, and weโ€™ll keep this value to one side. Now moving on to step two, weโ€™ll be using the sine rule again in triangle ๐ธ๐น๐บ to find the angle ๐น๐ธ๐บ, which is marked as a capital ๐ต on our diagram. And here in green we have the sine rule in the form thatโ€™s most useful for the side-side-angle criteria which weโ€™ll be using.

When applying the sine rule in this way, we look for a known side and angle pair, which we have in ๐‘Ž and capital ๐ด. We also have another known side, which is ๐‘, and an angle which weโ€™re interested in, which is capital ๐ต. Although you can apply the sine rule to any of the side angle pairs, here we wonโ€™t be interested in the ๐ถs, and so weโ€™ll get rid of this part of the formula.

Now as we did before, weโ€™ll sub in the three known values capital ๐ด, lowercase ๐‘Ž, and lowercase ๐‘. Weโ€™ll then work through the equation to solve and find uppercase ๐ต, which is the desired angle ๐น๐ธ๐บ. Substituting in, we find that sin of 70 degrees divided by 3.4634 is equal to sin of angle ๐น๐ธ๐บ divided by 2.5.

To simplify, we first multiply both sides of the equation by 2.5. And we find that 0.6782 et cetera is equal to sin of angle ๐น๐ธ๐บ. To continue, we then take arc sine of both sides of the equation. And we find that angle ๐น๐ธ๐บ is equal to 42.71 et cetera degrees. With stage two complete, weโ€™ll keep this value to one side and move on to stage three.

Weโ€™re now looking at the sum of angles in a triangle for the big triangle ๐ท๐ธ๐น. So here we know that the angle marked ๐น๐ท๐ธ added to angle ๐ท๐ธ๐น and ๐ธ๐น๐ท. Looking at the diagram, we can easily see that angle ๐ท๐ธ๐น is comprised of two smaller angles, one of which is the 30 degrees given by the question and the other is angle ๐น๐ธ๐บ that we have just found, which is 42.71 et cetera degrees. We also know angle ๐ธ๐น๐ท as it is given as 70 degrees in the diagram.

Letโ€™s sub these into the formula. So now that weโ€™ve substituted these values in, we want to solve for angle ๐น๐ท๐ธ. And we can do so by subtracting these two values from both sides of the equation. And performing this calculation, we find that angle ๐น๐ท๐ธ is equal to 37.28 dot dot dot degrees.

Weโ€™re now ready to move on to stage four of our method. Here we have our big triangle ๐ท๐ธ๐น drawn, with the relevant information labeled. And here we have a different form of the sine rule, which is relevant to the angle-angle-side situation that weโ€™ll be working with. Youโ€™ll know that weโ€™re directly ignoring ๐ถ this time as itโ€™s not relevant to our calculations.

And here weโ€™ve labeled our triangle with lowercase letters for the side lengths and uppercase letters for the angles. Weโ€™ve added a little dash to each of these symbols so as not to get confused with our earlier triangle labeling with ๐‘Žs, ๐‘s, and ๐‘s. And to keep things consistent, weโ€™ll add this onto our sine rule as well.

So here we should be able to see that weโ€™re aiming for side ๐ท๐ธ, which is represented by ๐‘Ž prime. And the other three terms are known, since we know angle ๐ด, angle ๐ต, and side length ๐‘. As weโ€™ve before, let sub everything in. We have ๐ท๐ธ divided by sin of 70 degrees is equal to 3.4 divided by sin of 37.28 dot dot dot degrees. By multiplying both sides of the equation by sin of 70 degrees, weโ€™re left with ๐ท๐ธ on its own on the left-hand side.

We then work through our calculations, and we find that ๐ท๐ธ is equal to 5.2747 centimeters. With this piece of the puzzle, weโ€™ve completed step four, and weโ€™re finally ready to work out the area of our big triangle.

For step five, weโ€™ll be using the area of a triangle formula that we mentioned earlier. Weโ€™ll also have the dashes to represent ๐‘Ž prime, ๐‘ prime, and capital ๐ถ prime to be consistent with the diagram on the right. And finally, weโ€™ll collect the relevant information.

We can observe that the angle at ๐ท๐ธ๐น is represented by ๐ถ prime. Based on the information weโ€™ve previously found, we can easily see that this angle ๐ท๐ธ๐น is the sum of 30 degrees and 42.71 degrees, which is 72.71 degrees. Labeling this on our small diagram on the right, we can see that the criterion of side-angle-side has been fulfilled, since we know side lengths ๐‘Ž prime, ๐‘ prime, and angle ๐ถ prime.

We perform one final set of substitutions with our known values. And working through the numbers, we find that the area of triangle ๐ท๐ธ๐น is equal to 8.5617 centimeters squared. And donโ€™t forget to add this squared to your unit since weโ€™re working with an area now, not a length.

As a final step, weโ€™ll note that our question asked for the answer to three significant figures. We can count our significant figures from left to right. Our answer of 8.561 will round down to 8.56, and this is our answer. The area of the triangle ๐ท๐ธ๐น, to three significant figures, is 8.56 centimeters squared.

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