WEBVTT
00:00:00.900 --> 00:00:05.860
In the triangle π΄π΅πΆ, one of the angles is the arithmetic mean of the other two.
00:00:06.160 --> 00:00:12.380
Find each angle of the triangle given the difference between the smaller and larger angles is 61 degrees.
00:00:12.680 --> 00:00:16.160
Itβs sensible to begin by defining each of the angles in our triangle.
00:00:16.280 --> 00:00:22.030
Weβll say that each of the three angles in our triangle are equal to π₯ degrees, π¦ degrees, and π§ degrees.
00:00:22.060 --> 00:00:28.000
Before even using any of the information in the question, we know that the sum of the angles is 180 degrees.
00:00:28.170 --> 00:00:32.310
So we can say that π₯ plus π¦ plus π§ must be equal to 180.
00:00:32.480 --> 00:00:40.320
If we then say that π₯ is the smallest and π§ is the largest, it follows that π¦ must be the arithmetic mean of π₯ and π§.
00:00:40.580 --> 00:00:47.340
In other words, when we find the sum of π₯ and π§ and then divide that by two, we end up with the third angle; thatβs π¦.
00:00:47.520 --> 00:00:49.850
So π₯ plus π§ over two equals π¦.
00:00:50.020 --> 00:00:52.990
Letβs neaten this up a little bit and multiply through by two.
00:00:53.170 --> 00:00:56.350
When we do, we find that π₯ plus π§ equals two π¦.
00:00:56.470 --> 00:01:01.540
The other piece of information we have is that the difference between the smaller and larger angles is 61 degrees.
00:01:01.600 --> 00:01:05.080
So π§ minus π₯ must be equal to 61.
00:01:05.980 --> 00:01:11.620
And weβve created a system of linear equations in three variables, π₯, π¦, and π§.
00:01:11.970 --> 00:01:14.060
So how do we solve each of these equations?
00:01:14.220 --> 00:01:18.030
Well, ideally, we want to create an equation in just one variable.
00:01:18.090 --> 00:01:22.790
And so, weβre gonna begin by manipulating the first two equations Iβve labelled one and two.
00:01:22.950 --> 00:01:29.700
If we subtract two π¦ from both sides of our second equation, we get π₯ minus two π¦ plus π§ equal zero.
00:01:29.760 --> 00:01:31.390
Letβs call this equation three.
00:01:31.820 --> 00:01:37.490
Notice that in equation one and equation three, we have exactly the same number of π₯s and π§s.
00:01:37.520 --> 00:01:40.390
The only thing that changes is the number of π¦s we have.
00:01:40.430 --> 00:01:44.400
So weβre going to subtract equation three from an equation one.
00:01:44.620 --> 00:01:47.450
This will have the effect of eliminating both π₯ and π§.
00:01:47.450 --> 00:01:49.880
And weβll just have an equation in terms of π¦.
00:01:49.940 --> 00:01:54.760
When we do, weβre left with π¦ minus negative two π¦ equals 180 minus zero.
00:01:54.860 --> 00:01:57.090
π¦ minus negative two π¦ is three π¦.
00:01:57.090 --> 00:02:00.450
And we can solve this equation for π¦ by dividing through by three.
00:02:00.570 --> 00:02:03.570
And when we do, we obtain π¦ to be equal to 60.
00:02:03.760 --> 00:02:06.510
So weβve calculated the sides of the middle angle.
00:02:06.510 --> 00:02:08.630
We need to work out the size of angle π₯ and π§.
00:02:08.910 --> 00:02:13.150
Weβre going to take the value of π¦ and substitute it into equation two.
00:02:13.370 --> 00:02:16.910
And when we do, weβll be left with two equations purely in terms of π₯ and π§.
00:02:17.300 --> 00:02:20.010
When we do, we get π₯ plus π§ equals two times 60.
00:02:20.010 --> 00:02:21.200
So thatβs 120.
00:02:21.300 --> 00:02:23.530
So weβve got equation four and five.
00:02:23.530 --> 00:02:25.320
These are purely in terms of π₯ and π§.
00:02:25.470 --> 00:02:29.070
Notice, though, that the signs of the π₯s are different in each equation.
00:02:29.280 --> 00:02:32.690
So weβre going to add these two equations, and that will eliminate the π₯.
00:02:32.720 --> 00:02:35.690
We could, alternatively, subtract the equations to eliminate the π§.
00:02:35.960 --> 00:02:41.600
When we add the expressions on the left-hand side, we get π§ minus π₯ plus π₯ plus π§, which is simply two π§.
00:02:41.600 --> 00:02:46.560
And on the right-hand side, we get 120 plus 61 which is 181.
00:02:46.770 --> 00:02:49.200
We solved this equation by dividing through by two.
00:02:49.350 --> 00:02:52.020
And we see that π§ is equal to 90.5.
00:02:52.300 --> 00:02:54.900
All thatβs left is to calculate the size of angle π₯.
00:02:54.950 --> 00:02:58.120
Itβs sensible to go back to either equation four or five.
00:02:58.340 --> 00:03:01.680
Now, Iβm going to choose equation five because there are no negatives.
00:03:01.880 --> 00:03:05.220
We get π₯ plus 90.5 equals 120.
00:03:05.280 --> 00:03:12.090
And then, we solve for π₯ by subtracting 90.5 from both sides, to find that π₯ is equal to 29.5.
00:03:12.380 --> 00:03:18.210
In ascending order, our angles are 29.5 degrees, 60 degrees, and 90.5 degrees.
00:03:18.400 --> 00:03:22.490
Now, we could absolutely check our working out by going back to equation one.
00:03:22.620 --> 00:03:27.460
We replace π₯ with 29.5, π¦ with 60, and π§ with 90.5.
00:03:27.620 --> 00:03:30.340
Their sum is 180 as required.
00:03:30.340 --> 00:03:32.000
So we know weβve done our working out correctly.
00:03:32.340 --> 00:03:36.760
The angles are 29.5, 60, and 90.5 degrees.