Triangle Geometry Problem: Find AD Length

Hey guys, today we're diving into a super interesting geometry problem involving a right-angled triangle. We're given a triangle ABC where angle A is a perfect 90 degrees and angle C is 15 degrees. We've also got point M, which is the midpoint of the hypotenuse BC, and point D, which is the projection of A onto BC. We know that the length of AM is $4 \sqrt{3}$ cm, and we need to figure out the length of AD. This is a classic type of problem that tests your understanding of triangle properties, so stick around and let's break it down step-by-step!

Understanding the Properties of Right-Angled Triangles

Alright, let's get straight into the nitty-gritty of this geometry math problem. We're dealing with a right-angled triangle ABC, which means one of its angles is 90 degrees. In our case, it's angle A. We're also given that angle C is 15 degrees. Now, the sum of angles in any triangle is always 180 degrees. So, if angle A is 90° and angle C is 15°, then angle B must be $180° - 90° - 15° = 75°$. This is a crucial piece of information we'll use later on. Another key thing to remember about right-angled triangles is the concept of the median to the hypotenuse. The problem states that M is the midpoint of BC, and AM is the median to the hypotenuse. A super important theorem here is that the median to the hypotenuse of a right-angled triangle is exactly half the length of the hypotenuse. This means $AM = BM = CM$. Since we're given that $AM = 4 \sqrt{3}$ cm, we now know that the entire hypotenuse BC is $2 \times AM = 2 \times 4 \sqrt{3} = 8 \sqrt{3}$ cm. This is a pretty neat shortcut, right? It simplifies things considerably. We're also told that D is the projection of A onto BC. This means that AD is perpendicular to BC, forming another right-angled triangle ADB and ADC. The line segment AD is also the altitude (or height) of the triangle ABC from vertex A to the hypotenuse BC. Finding the length of this altitude is often a key objective in these kinds of geometry math problems. We've established that $AM = 4 \sqrt{3}$ cm, and because M is the midpoint of BC, we also know that $BM = CM = 4 \sqrt{3}$ cm. Now, let's think about the triangle AMC. Since AM = CM, triangle AMC is an isosceles triangle. The angle MAC would be equal to angle MCA, which is 15 degrees. Similarly, in triangle AMB, since AM = BM, triangle AMB is also an isosceles triangle. The angle MAB would be equal to angle MBA, which is 75 degrees. This gives us a way to find the angles within triangle ABC: angle BAC = angle MAB + angle MAC = 75° + 15° = 90°, which matches the given information. So far, so good! The information is consistent, and we've unlocked some useful lengths and angles. Keep these in mind as we move forward to solve for AD.

Leveraging the Median to the Hypotenuse

So, guys, we've already established a couple of really cool facts about our right-angled triangle ABC. We know that $AM = 4 \sqrt{3}$ cm and since M is the midpoint of BC, AM is the median to the hypotenuse. Remember that awesome theorem? The median to the hypotenuse in a right-angled triangle is half the length of the hypotenuse. This means $AM = BM = CM = 4 \sqrt{3}$ cm. Consequently, the entire length of the hypotenuse BC is $2 \times AM = 8 \sqrt{3}$ cm. This is super helpful because it gives us the full length of the base we're working with. Now, let's shift our focus to the altitude AD. Remember, D is the point where A is projected onto BC, meaning AD is perpendicular to BC. We need to find the length of AD. We know the angles of triangle ABC are A = 90°, C = 15°, and B = 75°. In geometry math problems like this, there are a few ways to find the altitude. One common approach is to use trigonometry. We can use the area of the triangle. The area of triangle ABC can be calculated in two ways: (1) $\frac{1}{2} \times AB \times AC$ (using the two legs) or (2) $\frac{1}{2} \times BC \times AD$ (using the hypotenuse and the altitude to it). If we can find the lengths of AB and AC, we can calculate the area and then solve for AD. However, finding AB and AC directly might involve more steps. Let's consider using the properties of triangle ADC and ADB. Since AD is perpendicular to BC, both triangle ADC and triangle ADB are right-angled triangles. In triangle ADC, we have angle C = 15° and angle ADC = 90°. We want to find AD. We know CD. Hmm, we don't know CD yet. Let's look at triangle ADB. We have angle B = 75° and angle ADB = 90°. We want to find AD. We know BD. Again, we don't know BD yet. But we do know that $BD + CD = BC = 8 \sqrt{3}$ cm. Let's go back to the isosceles triangle AMC. We found that angle MAC = 15°. And in triangle AMB, angle MAB = 75°. This gives us a clear breakdown of angle BAC. Now, let's think about AD in relation to AM. We know AM = $4 \sqrt{3}$ cm. Consider the triangle ADC. We know angle C = 15° and angle ADC = 90°. In this right-angled triangle, $AD = AC \times \sin(15°)$. We also know $CD = AC \times \cos(15°)$. This doesn't seem to directly help us find AD without knowing AC. Let's try a different angle. What if we consider triangle AMB? We know angle B = 75° and angle AMB is an exterior angle to triangle AMC. Since angle MAC = 15°, the exterior angle AMB = angle MAC + angle MCA = 15° + 15° = 30°. This is a very useful piece of information! In triangle AMB, we have angles 75°, 30°, and angle BAM. The sum must be 180°, so angle BAM = $180° - 75° - 30° = 75°$. Wait, this means triangle AMB is isosceles with AM = BM, which we already knew! Okay, so the angle AMB = 30° is indeed correct. Now, let's use this angle. In triangle ADM, we have angle ADM = 90°. We know AM = $4 \sqrt{3}$ cm and angle AM D is the same as angle AMB, which is 30°. In right-angled triangle ADM, we can find AD using trigonometry. The side AD is opposite to the angle AMD (30°), and AM is the hypotenuse. So, $AD = AM \times \sin(30°)$. We know that $\sin(30°) = \frac{1}{2}$. Therefore, $AD = 4 \sqrt{3} \times \frac{1}{2} = 2 \sqrt{3}$ cm. Bingo! We found our answer using the properties of the median and trigonometry!

Trigonometric Approach and Final Calculation

Let's recap and solidify our understanding with a clear trigonometric approach to solve this geometry math problem. We have established that triangle ABC is a right-angled triangle with angle A = 90°, angle C = 15°, and consequently, angle B = 75°. Point M is the midpoint of the hypotenuse BC, and AM is the median. We are given that $AM = 4 \sqrt{3}$ cm. A fundamental property of right-angled triangles tells us that the median to the hypotenuse is half the length of the hypotenuse itself. This implies $AM = BM = CM = 4 \sqrt{3}$ cm. Therefore, the length of the hypotenuse BC is $2 \times AM = 8 \sqrt{3}$ cm. Our goal is to find the length of AD, where D is the projection of A onto BC, making AD the altitude. Consider the triangle AMC. Since $AM = CM$, it is an isosceles triangle. The angle MAC is equal to angle MCA, which is 15°. Now, let's look at the angles around point M on the line segment BC. The angle AMB is supplementary to angle AMC. However, it's more direct to consider triangle AMB. Since $AM = BM$, triangle AMB is also an isosceles triangle. The angle MAB is equal to angle MBA, which is 75°. Now, let's focus on the angle $\angle AMB$. This angle is an exterior angle to triangle AMC. The sum of the interior angles of triangle AMC is 180°. We know $\angle MAC = 15°$ and $\angle MCA = 15°$. So, $\angle AMC = 180° - (15° + 15°) = 180° - 30° = 150°$. Since angles AMB and AMC are supplementary (they form a straight line BC), $\angle AMB = 180° - \angle AMC = 180° - 150° = 30°$. Alternatively, as we noted before, the exterior angle $\angle AMB$ of triangle AMC is equal to the sum of the two opposite interior angles: $\angle AMB = \angle MAC + \angle MCA = 15° + 15° = 30°$. This is a very quick way to get this angle! Now, let's consider the right-angled triangle ADM. We know that $\angle ADM = 90°$ (because AD is the altitude), and we have just found that $\angle AMD = 30°$. The hypotenuse of this triangle is AM, with length $4 \sqrt{3}$ cm. We want to find the length of AD, which is the side opposite to the angle AMD. Using the sine function in triangle ADM: $\sin(\angle AMD) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AD}{AM}$. Plugging in the values we know: $\sin(30°) = \frac{AD}{4 \sqrt{3}}$. We know that $\sin(30°) = \frac{1}{2}$. So, the equation becomes: $\frac{1}{2} = \frac{AD}{4 \sqrt{3}}$. To solve for AD, we multiply both sides by $4 \sqrt{3}$: $AD = \frac{1}{2} \times 4 \sqrt{3} = 2 \sqrt{3}$ cm. This result matches option E. Therefore, the length of AD is $2 \sqrt{3}$ cm. This problem beautifully combines the properties of medians in right-angled triangles with basic trigonometry to arrive at the solution. Keep practicing these kinds of geometry math problems, guys, and you'll master them in no time!

The Answer

Based on our detailed calculations using the properties of medians and trigonometry, the length of AD is found to be $2 \sqrt{3}$ cm. This corresponds to option E among the choices provided.