Question Video: Finding the Missing Number in the Tangent Ratio for Angles in Equilateral Triangles

Video Transcript

Find the value of 𝐾 given 𝐴𝐵𝐶 is an equilateral triangle where point 𝐷 lies on 𝐴𝐵, 𝐴𝐷 equals five centimeters, 𝐷𝐵 equals 12 centimeters, and 𝐾 times the tangent of 𝑋 equals the square root of three.

We’ve been given an image to use here. And 𝐴𝐷 and 𝐵𝐷 have been labelled five and 12 respectively. We know that 𝐴𝐵𝐶 is an equilateral triangle. And that means the distance from 𝐶 to 𝐵 equals 17 centimeters. It equals five plus 12. 𝐴𝐶 also measures 17 centimeters. In equilateral triangles, every side has the same side length. There’s also something else we can say about this triangle. We know the measure of angle 𝐵 and angle 𝐴; they are 60 degrees.

In equilateral triangles, all three of the angles measure 60 degrees. The next thing we wanna know is that we’re dealing with a tangent ratio, but we’re not dealing with a right angle. And that means we’ll have to use a few different laws to solve this. The tangent of 𝑋 equals the sine of 𝑋 over the cosine of 𝑋, which means we’ll need to solve for both the sine and the cosine of this 𝑋 value. Angle 𝑋 sits within this smaller triangle inside our equilateral triangle.

We know two of the side lengths, side 12 and side 17, and we know an angle between these two side lengths: side-angle- side. We’re missing this length from 𝐶 to 𝐷. And because we have a side and angle and a side, we can use the law of cosine to find our third side length. Our 𝑏 squared our missing length is equal to 𝑎 squared plus 𝑐 squared minus two 𝑎𝑐 times the cosine of 𝐵. We use our other two side lengths 12 and 17 as 𝑎 and 𝑐: minus two times 12 times 17 times the cosine of angle 𝐵. Angle 𝐵 is equal to 60 degrees.

Working through it, one piece at a time, 12 squared equals 144, 17 squared equals 289, two times 12 times 17 equals 408, and cosine of 60 degrees is one-half. I can multiply 408 by half. That gives me 204. Copy everything else down: 144 plus 289 minus 204 equals 229. 𝑏 squared equals 229. We need to take the square root to find 𝑏 by itself.

𝑏 is equal to the square root of 229, and we’re actually not going to simplify that any further. We’ll just leave it as the square root of 229. In our yellow triangle, we now know all three of the sides: side-side-side. We can use these three sides to find the value of cosine of 𝑋. We’ll use the law of cosines again. In the law of cosines, if we want to find the angle value of 𝑋, we’ll use this 12 value on the outside.

It will be 12 squared equal to 17 squared plus the square root of 229 squared minus two times 17 times the square root of 229 times the cosine of 𝑋, which we don’t know. That’s our missing value. We need to solve for the cosine of 𝑋. Piece by piece, 12 squared equals 144. 17 squared equals 289. The square root of 229 squared equals 229. Two times 17 equals 34. And we’ll leave the square root of 229 there without simplifying it.

Bring down the cosine of 𝑋. Adding 289 and 229, we get 518. Bring everything else down. From there, we subtract 518 from both sides of the equation, cancels out on the right. On the left, we have negative 374. And then, we’ll bring everything else down. To get the cosine of 𝑋 by itself, I’ll divide by negative 34 times the square root of 229 on both sides. On the right it cancels out, the cosine of 𝑋 is equal to negative 374 over negative 34 times the square root of 229.

At this point, you might be wondering where on earth are we going, but stick with me! What I’m gonna do now is we’re gonna divide negative 374 by negative 34. That actually equals positive 11. And we are not going to simplify the square root of 229. We just found that the cosine of 𝑋 equals 11 over the square root of 229. And now we need the sine of 𝑋. To find that, we’ll use the law of sine that tells us the sine of angle 𝐵 over its opposite length 𝑏 is equal to the sine of angle 𝐴 over its opposite length 𝑎.

We use the sine of an angle and its opposite length, the sine of an angle and its opposite length. We have the sine of 60 degrees over the square root of 229, sine of 𝑋, which we don’t know over 12. To solve for sine of 𝑋, I’m going to multiply the right side by 12 over one. And if I multiply by 12 over one on the right, I need to do it on the left. Again! Stay with me just a little bit longer and it’s all going to start coming together!

What is the sine of 60 degrees? Sine of 60 degrees is the square root of three over two. Okay, I’ll just give us a little bit more room. Two times the square root of three over two equals six times the square root of three. And again, we just keep bringing along the square root of 229. And this is our sine of 𝑋. If we use the cosine and the sine to find the tangent, we only need their numerators because their denominators are going to cancel out anyway.

We would put the numerator of the sine of 𝑋 and the numerator of the cosine of 𝑋 in for tangent. Tangent of 𝑋 equals six times the square root of three over 11. We’re given this other piece of information: that 𝐾 times the tangent of 𝑋 equals the square root of three. And our question is wondering what the value of 𝐾 is. We’re going to take what we found tangent of 𝑋 be equal to and plug it in.

𝐾 times six times the square root of three over 11 is equal to the square root of three. I’m just going to break the factors up so we can see them a little bit more clearly. If I divide by the square root of three on both sides, on the left it goes away and, on the right, the square root of three over the square root of three equals one. 𝐾 times six over 11 equals one. Then we multiply by 11 over six on both sides. On the left, they cancel out, leaving us with 𝐾 being equal to one times 11 over six. 𝐾 equals 11 over six. And that’s how you do it! Walk through it step-by-step, finding each of the pieces, and finally we find that 𝐾 equals 11 over six.