Question Video: Finding the Unknown Length in a Right Triangle Using Trigonometry Where the Unknown Is on the Bottom of the Fraction | Nagwa Question Video: Finding the Unknown Length in a Right Triangle Using Trigonometry Where the Unknown Is on the Bottom of the Fraction | Nagwa

Question Video: Finding the Unknown Length in a Right Triangle Using Trigonometry Where the Unknown Is on the Bottom of the Fraction Mathematics • First Year of Secondary School

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Find 𝑥 to two decimal places.

03:51

Video Transcript

Find 𝑥 to two decimal places.

What are some things we know about this problem, simply by looking at the picture? We know that we’re dealing with a triangle. More specifically, this is a right triangle. We know that it’s a right triangle because of this symbol on our picture. We know that 𝑥 is the hypotenuse, the longest side. We know that because the 𝑥 is the side that’s opposite the right angle. We’re given an angle measure of 20 degrees. We’re also given the length of the side that is opposite the 20-degree angle.

From here, we’re going to need some information that we don’t have simply by looking at the picture. This is where trigonometry functions come in. The sine of some angle equals the opposite side length over the hypotenuse side length. The cosine of an angle equals the adjacent side length over the hypotenuse side length. And the tangent would be the opposite side length over the adjacent side length. We’re given an angle measure. We’re missing the hypotenuse; we wanna know what that value is. And we’re working with the side length that’s opposite the angle that we’ve been given.

Since we have an opposite side length and the hypotenuse, we should use the sine function. If we plug in our information into a sine function, it looks like this: sine of 20 degrees equals 12 over 𝑥. But remember, we’re solving for 𝑥, so we need to get 𝑥 by itself. To get 𝑥 by itself, we’ll first need to get it out of the denominator here. We can multiply both sides of our equation by 𝑥 over one. On the right-hand side, we’re left with 12 because the 𝑥s cancel out. On the left-hand side, we’ll have 𝑥 times sine of 20 degrees. 𝑥 is still not by itself, so we have one more step. We can divide both sides of our equation by sine of 20 degrees. On the left side, sine of 20 degrees divided by sine of 20 degrees equals one; they cancel out. 𝑥 is the only thing remaining.

What we have now is something that says the side length 𝑥 will be equal to 12 divided by the sine of 20 degrees. If you plug that into your calculator, you should get something like this: 35.0856528. But our question is asking us to find the length of 𝑥 to two decimal places. We can look at the third decimal place to see if our eight will round up or stay the same. Because of the five, the eight will round up leaving us with 𝑥 being approximately 35.09. We weren’t given what units, so we can say 35.09 units.

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