# Question Video: Using Trigonometry to Solve Problems Involving Isosceles Trapezoids Mathematics • 10th Grade

Find (17 tan 𝐵 cos 𝐶)/(sin² 𝐶 + cos² 𝐵), given that 𝐴𝐵𝐶𝐷 is an isosceles trapezoid, where line segment 𝐴𝐷 ‖ line segment 𝐵𝐶, 𝐴𝐷 = 8 cm, 𝐴𝐵 = 17 cm, and 𝐵𝐶 = 24 cm.

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### Video Transcript

Find 17 tan of 𝐵 cos of 𝐶 over sin squared of 𝐶 plus cos squared of 𝐵, given that 𝐴𝐵𝐶𝐷 is an isosceles trapezoid, where line segment 𝐴𝐷 is parallel to line segment 𝐵𝐶, 𝐴𝐷 equals eight centimeters, 𝐴𝐵 equals 17 centimeters, and 𝐵𝐶 equals 24 centimeters.

And then we’ve been given a diagram of the isosceles trapezoid. So let’s first remind ourselves what we mean when we talk about an isosceles trapezoid. Well, in an isosceles trapezoid, the two nonparallel sides are equal in length. This also means that the angles that these nonparallel sides make with the two parallel sides are equal.

Now, in this question, we’re looking to find an expression for various trigonometric functions of angles 𝐵 and 𝐶. So let’s define angles 𝐵 and 𝐶 to both be equal to 𝜃. We know the trigonometric ratios for a right triangle with included angle 𝜃 tell us that sin of 𝜃 is equal to opposite over hypotenuse. cos of 𝜃 is equal to adjacent divided by the hypotenuse. And tan of 𝜃 is equal to opposite divided by adjacent. So where are we going to find a right triangle in our diagram? Well, we can add a perpendicular directly up from point 𝐴. We know that this triangle has an included angle of 𝜃, and it has a hypotenuse of 17 centimeters. So let’s work out the length of the side adjacent to the angle 𝜃, and we’ll call that 𝑥 centimeters.

Since the trapezoid is isosceles, we know if we add a second perpendicular directly up from point 𝐷, we end up with another triangle with an adjacent side of 𝑥 centimeters. Then, we also have a rectangle, so the remaining portion of line segment 𝐵𝐶 must be eight centimeters in length. We can therefore say that since line segment 𝐵𝐶 consists of three portions, 𝑥, eight, and 𝑥, which sum to make 24 centimeters, two 𝑥 plus eight must be equal to 24. We now have an equation we can solve for 𝑥. We begin by subtracting eight from both sides, so two 𝑥 equals 16. Then, dividing through by two, we find 𝑥 is equal to eight.

And so we now have a right triangle for which we know two of the dimensions. We are going to work out the third dimension since at some point we’re going to calculate sin of 𝜃, cos of 𝜃, and tan of 𝜃. Here is the triangle we’re interested in. We can use the Pythagorean theorem to work out the missing length in this triangle, the side opposite the included angle. We might alternatively notice that we have a Pythagorean triple, but let’s use the Pythagorean theorem to check.

Let’s call the length of the side that we’re trying to find 𝑦 centimeters. Then, by the Pythagorean theorem, the sum of the squares of the two shorter sides must be equal to the square of the hypotenuse. So eight squared plus 𝑦 squared is equal to 17 squared. 64 then plus 𝑦 squared equals 289. By subtracting 64 from both sides, we find that 𝑦 squared is equal to 225. Then, taking the positive square root of 225, we find that 𝑦 is equal to 15. And so we now have the third side of the triangle. And there is a Pythagorean triple we might recognize. Eight squared plus 15 squared is 17 squared.

We now have everything we need to calculate the value of tan of 𝐵, cos of 𝐶, sin of 𝐶, and cos of 𝐵. tan of 𝐵 is of course tan 𝜃, and we know that tan 𝜃 is opposite over adjacent. In our triangle, the opposite side is 15 centimeters and the adjacent is eight. So tan of 𝜃 is 15 over eight. Next, we need to find the value of cos of 𝐶 and cos of 𝐵. But since both 𝐶 and 𝐵 are 𝜃, we simply need to find the value of cos of 𝜃. Now, of course, that’s adjacent divided by hypotenuse. So it’s eight over 17. cos of 𝜃 is eight seventeenths. Finally, we need to know the value of sin of 𝐶. So that’s sin of 𝜃, where sin of 𝜃 is opposite over hypotenuse. So sin of 𝜃 is 15 over 17.

Let’s now find the value of 17 tan of 𝐵 cos of 𝐶 over sin squared of 𝐶 plus cos squared of 𝐵. The numerator is 17 times tan of 𝐵, 15 over eight, times cos of 𝐶, eight over 17. Then, the denominator, sin squared of 𝐶 plus cos squared of 𝐵, is eight seventeenths squared plus 15 over 17 squared.

Now, before we try and evaluate this in our head, we might notice that we can simplify somewhat. First, on our numerator, we can simplify by dividing through by 17. Similarly, we can divide through by eight, and that leaves simply 15 on the numerator of our expression. Next, remember that eight squared plus 15 squared is 17 squared. That’s our Pythagorean triple. So the denominator of this expression is actually going to equal one. This means we get 15 over one, which of course is simply equal to 15.

So the value of 17 tan of 𝐵 cos of 𝐶 over sin squared of 𝐶 plus cos squared of 𝐵 is 15.