The 5 equal lengths that make up the two sides of the largest triangle tell us that we are dealing with 5 similar triangles. The largest triangle has sides 15:25:30, and the sides of all 5 triangles will have an equivalent ratio. Reduced, the ratio is 3:5:6, which happens to be the dimensions of the smallest triangle. We want to find the length of BD, the base of a triangle with sides of 6 and 10. This is twice as big as the smallest triangle, so the base BD must be 6 × 2 =12, which is (C).

Use SOHCAHTOA and your calculator to find the height of the flagpole. From the 70° angle, you know the adjacent side of the triangle, and you want to find the opposite side, so you need to use tangent. Tangent ​\( \frac{opposite}{adjacent} \)​ , so tan 70° = ​\( \frac{x}{10m} \)​, where x is the height of the flagpole up to the ball. Isolate x by multiplying both sides by 10: 10 tan 70° = x. Use your calculator to find that 10 tan 70° = 27.47, which is closest to (C).

Don’t forget that you can plug in numbers on geometry questions. Let’s make b = 70° and a = 30°. So the third angle in the triangle is 80°. You know that c would be 80°, because it is opposite an 80° angle. Your target answer is a = 30°, so plug in 80° and 70° to find it. The only possible answer is (D).

There’s a lot going on in this problem! But if we take it piece by piece, we’ll crack it. Let’s start filling in some information. The first thing the problem tells us is that triangle ABC is equilateral. Mark 60 degree angles on the figure. Next, we see that angle AEF is a right angle. Write that in as well. The problem also conveniently tells us that D and F are the midpoints of AB and AC, respectively. Therefore, AD and AF are 2. Finally, the last piece of information reveals that E is the midpoint of DF; mark DE and EF as equal. Triangle AEF is a right triangle, with a hypotenuse of 2 and a leg of 1. Hmm, perhaps the good ol’ Pythagorean theorem can help us. Plug the numbers into the theorem, and you’ll see that the answer is (B).