WEBVTT
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π΄π΅πΆ is a triangle, where angle π΄ is 70 degrees, π΅πΆ is three centimeters, and π΄πΆ is 39 centimeters.
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If the triangle exists, find all the possible values for the other lengths and angles in triangle π΄π΅πΆ giving the lengths to two decimal places and angles to the nearest degree.
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Letβs start by drawing a sketch of what this triangle would look like based on the information given.
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Weβre asked that if this triangle exists to find the other lengths and angles.
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When solving for missing sides or angles in a triangle like this, we often use the sine rule and the cosine rule.
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But in this case, because weβve been given two sides and a nonincluded angle, weβre going to use the sine rule.
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Remember that this is sin π΄ over π equals sin π΅ over π equals sin πΆ over π, where the uppercase π΄π΅πΆ are the angles at the respective points and the lowercase πππ represent the opposite side to the respective angle.
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So in this case, we have angle π΄ is 70 degrees.
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So this side is lowercase π, so lowercase π equals three.
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And we also have that this side is side π and thatβs 39.
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So because weβve got angle π΄ and side π and side π, letβs try to find angle π΅ first of all.
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So weβre going to use the fact that sin π΄ over π equals sin π΅ over π.
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Letβs fill in the bits that we know.
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That gives us sin 70 over three equals sin π΅ over 39.
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And by rearrangement, this gives us that sin of π΅ equals 39 multiplied by sin 70 over three.
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And we can calculate sin of π΅ to be 12.22 to two decimal places.
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However, weβve just encountered a little bit of a problem here.
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The value of sin can never be greater than one.
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And if we were to try and go ahead anyway and find the value of angle π΅, by doing the inverse sin of 12.22, we would find ourselves with an error message.
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And this is demonstrated in the fact that sin of an angle equals the opposite side over the hypotenuse.
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And the hypotenuse is always the longest length of a right-angled triangle.
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So the hypotenuse is always going to be greater than the opposite length.
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So because the hypotenuse is bigger than the opposite, the opposite over the hypotenuse cannot be greater than one.
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And actually, we see here that if we draw a straight line down from angle πΆ to the length π΄π΅ and since we know that sin of an angle is the opposite over the hypotenuse, using the right-angled triangle on the left, we have that sin of 70 equals the opposite over 39.
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And we can rearrange this and find that the opposite length is 36.65 centimeters to two decimal places.
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So if this line is indeed 36.65 centimeters, then the right-angled triangle to the right has a hypotenuse of three and the opposite side is 36.65.
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But that canβt possibly work because we know the hypotenuse is always the longest side in a right-angled triangle.
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So the diagram is not possible, and therefore this triangle does not exist.