# Question Video: Solving a Triangle given Two Sides' Lengths and an Angle Mathematics

𝐴𝐵𝐶 is a triangle, where 𝑚∠𝐴 = 70°, 𝐵𝐶 = 3 cm, and 𝐴𝐶 = 39 cm. If the triangle exists, find all the possible values for the other lengths and angles in △𝐴𝐵𝐶 giving the lengths to two decimal places and angles to the nearest degree.

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### Video Transcript

𝐴𝐵𝐶 is a triangle, where angle 𝐴 is 70 degrees, 𝐵𝐶 is three centimeters, and 𝐴𝐶 is 39 centimeters. If the triangle exists, find all the possible values for the other lengths and angles in triangle 𝐴𝐵𝐶 giving the lengths to two decimal places and angles to the nearest degree.

Let’s start by drawing a sketch of what this triangle would look like based on the information given. We’re asked that if this triangle exists to find the other lengths and angles. When solving for missing sides or angles in a triangle like this, we often use the sine rule and the cosine rule. But in this case, because we’ve been given two sides and a nonincluded angle, we’re going to use the sine rule. Remember that this is sin 𝐴 over 𝑎 equals sin 𝐵 over 𝑏 equals sin 𝐶 over 𝑐, where the uppercase 𝐴𝐵𝐶 are the angles at the respective points and the lowercase 𝑎𝑏𝑐 represent the opposite side to the respective angle.

So in this case, we have angle 𝐴 is 70 degrees. So this side is lowercase 𝑎, so lowercase 𝑎 equals three. And we also have that this side is side 𝑏 and that’s 39. So because we’ve got angle 𝐴 and side 𝑎 and side 𝑏, let’s try to find angle 𝐵 first of all. So we’re going to use the fact that sin 𝐴 over 𝑎 equals sin 𝐵 over 𝑏. Let’s fill in the bits that we know. That gives us sin 70 over three equals sin 𝐵 over 39. And by rearrangement, this gives us that sin of 𝐵 equals 39 multiplied by sin 70 over three. And we can calculate sin of 𝐵 to be 12.22 to two decimal places.

However, we’ve just encountered a little bit of a problem here. The value of sin can never be greater than one. And if we were to try and go ahead anyway and find the value of angle 𝐵, by doing the inverse sin of 12.22, we would find ourselves with an error message. And this is demonstrated in the fact that sin of an angle equals the opposite side over the hypotenuse. And the hypotenuse is always the longest length of a right-angled triangle. So the hypotenuse is always going to be greater than the opposite length. So because the hypotenuse is bigger than the opposite, the opposite over the hypotenuse cannot be greater than one.

And actually, we see here that if we draw a straight line down from angle 𝐶 to the length 𝐴𝐵 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. And we can rearrange this and find that the opposite length is 36.65 centimeters to two decimal places. 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. But that can’t possibly work because we know the hypotenuse is always the longest side in a right-angled triangle. So the diagram is not possible, and therefore this triangle does not exist.

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