![]() The leg opposite angle X is x, the leg adjacent to angle X is y, and the hypotenuse is z, so: In triangle XYZ shown, what are sin X, cos X and tan X? TANGENT is OPPOSITE LEG divided by ADJACENT LEG. SINE is OPPOSITE LEG divided by HYPOTENUSE.ĬOSINE is ADJACENT LEG divided by HYPOTENUSE. How do we remember that? Well, when I was in high school, we were given the following mnemonic: SOHCAHTOA (pronounce it saw-ca-toe-a). In summary, our basic trig functions are: That is referred to as the tangent of A, and we abbreviate that as: tan A. Then there's the ratio of the opposite leg to the adjacent leg: a/b. ![]() We've talked about the ratio of the opposite leg to the hypotenuse - what about the ratio of the adjacent leg to the hypotenuse? We'll call that ratio the cosine of A, and we'll abbreviate it cos A. For now we'll just worry about the acute angles. So we're going to leave that alone for now. What about sin C? That's a funny one, because there is no leg opposite angle C! Angle C is 90 degrees, and the side opposite is the hypotenuse, not a leg! So our definition doesn't exactly work out. So what is sin B? Well, it's the length of the side opposite angle B, divided by the length of the hypotenuse. How do we calculate sin A? It's the length of the leg opposite angle A, divided by the length of the hypotenuse. This is an important ratio here, this "leg over hypotenuse" ratio. Let's put that into words: in any two similar right triangles, the ratio of a leg to the hypotenuse of one triangle is equal to the ratio of the corresponding leg to the hypotenuse of the other triangle. In the equation above, multiply both sides by x, and divide both sides by c. Since they are similar, we know that the lengths of corresponding sides are in the same ratio. The similarity statement for these two triangles is: They are not congruent, however, if I tell you that angle A is equal to angle X, that's enough to conclude that they are similar. DE is 2 and 2/5.These are two right triangles with right angles at C and Z. Here- what we actually have to figure out-Ħ and 2/5, minus 4, minus CD right over here. Length- CE right over here- this is 6 and 2/5. Ratio of CB over CA is going to be equal to That the ratio between CB to CA- so let's Of corresponding sides are going to be constant. To triangle CAE, which means that the ratio Triangle CBD is similar- not congruent- it is similar Write it in the right order when you write your similarity. And once again, this isĪn important thing to do, is to make sure that you The corresponding angles, are congruent to each other. Stopped at two angles, but we've actually shown thatĪll three angles of these two triangles, all three of Triangles- so I'm looking at triangle CBDĪnd triangle CAE- they both share this angle up here. Once again, correspondingĪngles for transversal. We also know that thisĪngle right over here is going to be congruent to To be congruent to that angle because you could view And so we know correspondingĪngles are congruent. Let me draw a littleĭifferent problem now. ![]() Knowing that the ratio between the corresponding And then we get CE isĮqual to 12 over 5, which is the same thing 5 times the length of CE isĮqual to 3 times 4, which is just going to be equal to 12. Is really just multiplying both sides by both denominators. To be equal to- what's the corresponding side to CE? The correspondingĮqual to CA over CE. And I'm using BC and DCīecause we know those values. Of BC over DC right over here is going to be equal to ![]() The way that we've written down the similarity. The corresponding side for BC is going to be DC. Ratio of corresponding sides are going to be the same. Now, what does that do for us? Well, that tells us that the Then, vertex B right over here corresponds to vertex D. Is similar to triangle- so this vertex A corresponds Your, I guess, your ratios or so that you do know And that's really important-Ĭorrespond to what side so that you don't mess up It so that we have the same corresponding vertices. To say that they are similar, even before doing that. This angle and this angle are also congruent byĪlternate interior angles, but we don't have to. We have two triangles and two of the correspondingĪngles are the same. Or you could say that, if youĬontinue this transversal, you would have a correspondingĪngle with CDE right up here and that this one'sĪngle and this angle are going to be congruent. Might jump out at you is that angle CDE is anĪlternate interior angle with CBA. Might jump out at you is that this angle and thisĪngle are vertical angles. And then, we have these twoĮssentially transversals that form these two triangles. Over here, we're asked to find out the length
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