WEBVTT
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π΄π΅, πΆπ·, and πΈπΉ are straight lines.
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π΄π΅ is parallel to πΆπ·.
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Find the value of π¦.
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We can see that π¦ has been used in an expression for the size of one of the angles in this diagram.
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We were told in the question that the line π΄π΅ is parallel to the line πΆπ·.
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So in order to answer this question, weβre going to need to use some facts about angles in parallel lines.
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Notice that the expressions weβve been given for two other angles in this diagram are in terms of a second variable π₯.
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So weβre probably going to need to calculate the value of π₯ before we can calculate the value of π¦.
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Iβm going to introduce the letters πΊ and π» onto the diagram to represent the points where the line πΈπΉ crosses π΄π΅ and πΆπ·.
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Letβs look first of all at the angles π΄πΊπ» and πΆπ»πΈ.
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And we notice that theyβre enclosed within an πΉ shape.
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Itβs backwards.
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But itβs still an πΉ shape.
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The proper term for angles which are enclosed within an πΉ shape is corresponding angles.
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So we can conclude that angle π΄πΊπ» and πΆπ»πΈ are corresponding angles.
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Corresponding angles are equal to each other.
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So angle π΄πΊπ» is equal to angle πΆπ»πΈ.
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This means that we can add the expression of two π₯ plus 27 degrees to the top part of our diagram.
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Now letβs look at angles π΄πΊπΉ and π΄πΊπ».
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Thatβs the angle I have marked in pink and the angle I have marked in orange.
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These two angles lie on a straight line.
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And we know that the sum of angles on a straight line is 180 degrees.
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So angle π΄πΊπΉ plus angle π΄πΊπ» equals 180 degrees.
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We can form an equation by adding the expressions for these two angles and setting it equal to 180.
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We have four π₯ plus three plus two π₯ plus 27 equals 180.
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We can simplify this equation by grouping like terms.
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Firstly, four π₯ plus two π₯ is six π₯ and three plus 27 is 30.
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We are now in a position to be able to solve this equation.
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The first step is to subtract 30 from each side of the equation.
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On the left-hand side, weβre left with six π₯ and on the right-hand side, 180 minus 30 is 150.
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Next, we need to divide both sides of this equation by six.
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On the left-hand side six π₯ divided by six is just π₯ and on the right-hand side, 150 divided by six is 25.
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You can work this out with a bit of logic if you remember that four lots of 25 are 100.
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So six lots of 25 are 150 or you can use a short division method.
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There are no sixes is in one.
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So we carry the one into the next column.
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There are two sixes in 15 as two sixes are 12 with a remainder of three.
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And there are five sixes in 30 with no remainder, giving our answer of 25.
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So weβve found the value of π₯.
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But how would this help us with the question which was to find the value of π¦?
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There are two ways that we could now work out the value of π¦.
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Firstly, we could note that angle π΄πΊπ» and π΅πΊπ» are on a straight line, which means the sum of these two angles must be 180 degrees.
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As we know the value of π₯, we can work out the size of angle π΄πΊπ» and then form an equation to find the value of π¦.
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Or we can use the fact that angles π΄πΊπΉ and π΅πΊπ» are vertically opposite angles.
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Theyβre formed by the intersection of two straight lines.
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And we know that vertically opposite angles are equal.
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Letβs use this method.
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Angle π΄πΊπΉ is four π₯ plus three degrees.
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And we know the value of π₯ is 25.
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So we can work out the size of this angle by substituting 25 for π₯.
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Four multiplied by 25 is 100 and adding three gives 103.
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So angle π΄πΊπΉ is 103 degrees.
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But as vertically opposite angles are equal, angle π΅πΊπ» is also equal to 103 degrees.
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So we can form an equation: five π¦ minus two equals 103.
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To solve for π¦, we must first add two to each side, giving five π¦ equals 105, and then divide both sides of the equation by five to give π¦ equals 21.
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You can see that 105 divided by five is 21, either using a short division method or remembering that five times 20 is 100.
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So five times 21 will be 105.
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We found the value of π¦.
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π¦ is equal to 21.