Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 3 β€’ Question 6

The two right-angled triangles shown in the diagram are similar. a) Write down the value of tan πœƒ, give your answer as a fraction in its simplest form. b) Hence, or otherwise, work out the value of π‘₯.

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Video Transcript

The two right-angled triangles shown in the diagram are similar. Part a wants us to write down the value of tangent of πœƒ, giving our answer as a fraction in its simplest form. In part b, hence or otherwise, work out the value of π‘₯.

In order to find out the value of tangent πœƒ, we need to remember our trig ratios. The phrase SOH CAH TOA reminds us of the relationship of sine, cosine, and tangent in right triangles. In this problem, we’re only interested in the tangent ratio. The tangent of πœƒ equals the opposite side length over the adjacent side length. Before we go any further, we need to know a property of similar triangles. In similar triangles, corresponding angles are equal. This means that angle πœƒ has the same measure in both triangle 𝐴𝐡𝐢 and triangle 𝐸𝐷𝐢. It also means that tangent πœƒ is the same in both the large and small triangle.

In the larger triangle, triangle 𝐴𝐡𝐢, we have an opposite side length of 12 centimeters. The numerator of the tangent of πœƒ is then 12. The adjacent side length in the larger triangle is nine. That makes our denominator nine. We could say that the tangent of πœƒ equals twelve-ninths, but our instruction explicitly asked for the answer as a fraction in its simplest form. We can reduce twelve-ninths by dividing the numerator and the denominator by three. 12 divided by three is four. Nine divided by three is three. The tangent of πœƒ equals four-thirds.

Moving on to part b, we have two options: hence or otherwise. The hence option means use the information from part a to solve part b. Using tangent of πœƒ equals four-thirds, we’ll find the tangent ratio of the smaller triangle, the opposite eight centimeters over the adjacent π‘₯ centimeters. We’ve already said that these tangent ratios are equal to each other. Four- thirds is equal to eight over π‘₯. To go from four to eight, we multiply by two. And if we multiply the numerator by two, to keep it equivalent, we multiply the denominator by two. Three times two equals π‘₯, then π‘₯ is six. The missing side length is six centimeters.

We can also find the length of 𝐢𝐷 without relying on the tangent ratio. This time we’ll use the scale factor. When two triangles are similar, you can take a side length of the smaller triangle, multiply it by the scale factor, and that will equal the corresponding side length in the larger triangle. If we multiplied side then 𝐷𝐸, that’s eight centimeters, by the scale factor, we would get 12 centimeters. To find the scale factor, we divide both sides of this equation by eight. The scale factor is 12 divided by eight, which is one and a half. Eight times one and a half equals 12.

Now we can apply the scale factor to solve for π‘₯. We’re missing a side length in the smaller triangle. We know the scale factor is one and a half. And its corresponding side length is nine centimeters in the larger triangle. To solve for π‘₯, we divide both sides of the equation by one and a half. Nine divided by one and a half equals six. Again, we see that the missing side length measured six centimeters.

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