Right Triangle Trigonometry

The answer is A. As always, draw the triangle. If you remember that the hypotenuse (longest side) of the triangle must be across from the right angle, that will help you draw the figure more quickly. Alternatively, you can note that side $DE$ must be across from angle F, and “across” is precisely what we are seeking to find sine (opposite/hypotenuse). Whether you reason this way or use the diagram, the opposite is 14 and the hypotenuse must be $50$. $\frac{14}{50}$ reduces to $\frac{7}{25}$ when you divide by two.

One final note: it’s always worth noting Pythagorean triples when they appear, and we have one here: $7:24:25$ is one of the less common triples, but a good one to know nevertheless!

The answer is B. The key to this question is recognizing the $30-60-90$ special right triangle implied by the angles given. $\angle C$ must be the right angle, so we know that $\angle B$ is the smallest angle. This means (as shown in our lesson on triangles) that it must be across from the smallest side. As shown in the CLT reference list (though hopefully also solidly memorized by you!), the smallest side is represented by the $1$ in the $1:\sqrt{3}:2$ ratio. This means that the ratio of the opposite in this case (smallest side) to the adjacent (middle side) is $1:\sqrt{3}$. Since opposite/adjacent is what we need for tangent, we are close to our answer.

What remains to be done? The rules of algebra tell us that a radical (in this case, a square root) in the denominator is not considered simplified, so we need to rationalize the denominator by multiplying both top and bottom of the fraction by that denominator. That gives us the following:

$$\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{(\sqrt{3})(\sqrt{3})}=\frac{\sqrt{3}}{3}$$

The answer is C. There are a couple of pieces of hidden, implied information here. We can’t yet use the area of $8$ because we don’t know the base or height of the triangle. However, the fact that the tangent of the triangle is $1$ is useful. This means that $\frac{opp}{adj}=1$; multiplying both sides by $adj$, we learn that $opp=adj$. What does this tell us? We have an isosceles triangle, also known as a $45-45-90$.

How do we move forward with this information toward the cosine of the triangle? The special right triangle in this case must have a ratio of $1:1:\sqrt{2}$, so that’s it! The cosine of that only one acute angle but both acute triangles must be $\frac{1}{\sqrt{2}}$ ($\frac{adj}{hyp}$). As seen previously, we need to multiple both parts of the fraction by $\sqrt{2}$, and the result is choice $C$.

Note that we didn’t ever use the area of the triangle! This is what’s known as extraneous (unnecessary) information in a problem, and identifying when you don’t need to use certain information can be important for efficient work on a standardized test like the CLT. However, just for the extra practice, can you use the area of this triangle to find out the lengths of its sides?

The sides are $4$, $4$, and $4\sqrt{2}$. Because we know that the two legs here are congruent and that, in a right triangle, the length of the legs is equal to the base and height, we can use the area formula to posit that $(\frac{1}{2})(x)(x)=8$ in this case. Solving for $x$, we get $4$, and the $45-45-90$ ratio tells us that the hypotenuse is the leg length of $4\times \sqrt{2}$.

The answer is A. We might copy the figure and note the sine ratio here. However, since we (must!) note that trig ratios are only ratios, instead of labeling two sides $8$ and $17$, let’s label them $8x$ and $17x$ to show that what we have is a ratio rather than fixed values. Do you remember the Pythagorean triple here? It is $8:15:17$, so we can label the other leg $15x$.

The skill we just demonstrated can be helpful in many problems, so it’s worth considering, but in this case we don’t end up needing it. The ratio of $8:17$ shows us that we already have an example of what the student describes: $8$ is less than half of $17$. We need to find an example of a right triangle where both legs have lengths more than half of that of the hypotenuse.

Jumping to the wrong answers first, we can note that choice $C$ actually poses the same triangle we already have: the $8:15:17$ triangle. So that choice doesn’t provide any extra information. Choice $B$ involves a common Pythagorean triple: $5:12:13$. $5$ is certainly less than half of $13$, so this answer does not provide a counterexample.

Choice $D$ might at first seem like the answer because $24$ is way more than half of $25$, so perhaps the other leg of the triangle is also more than half of $25$. But no such luck: the other side here, given by a lesser-known Pythagorean triangle, is $7$. (If you don’t know that triple, you’ll have to do some Pythagorean theorem calculations with rather long numbers. $24^2$ is $576$ and $25^2$ is $625$. $625-576=49$, and the square root of $49$ is $7$. You can see what a valuable shortcut a Pythagorean triple can provide!)

This leaves answer $A$: a version of the classic $3:4:5$ triangle. Did you recognize it as a $3:4:5$ example? Since $20$ is $4\times 5$ and $16$ is $4\times 4$, the remaining side must be $4\times 3=12$. $12$ and $16$ are both more than half of $20$, so this is our counterexample. If necessary, work on recognizing your triples, because if you don’t, you have to do an involved calculation with the Pythagorean theorem to find the third side of $12$.

The answer is B. If you didn’t already do so in attempting this problem, take a stab at drawing a picture of this situation. As true in so many other situations, visualizing it clearly is key. Now compare your figure to ours, below.