Solving For Sin2α In Triangle ABC: A Geometry Challenge

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Hey guys! Let's dive into a cool geometry problem today that involves finding the value of sin⁡2α\sin 2\alpha within a triangle. This problem combines several geometric principles and trigonometric relationships, making it a fantastic exercise for anyone looking to sharpen their math skills. So, grab your pencils, and let’s get started!

Understanding the Problem

So, here’s the setup: We've got a triangle ABC. In this triangle, we know that twice the measure of angle BAD is equal to the measure of angle ACB, and both are equal to 2α2\alpha. That’s our key angle right there! We also know that AD is perpendicular to BC, which gives us a right angle – super helpful. And, we’re given the lengths of AC (10 cm) and BD (4 cm). The mission? Find the value of sin⁡2α\sin 2\alpha. The options we have are: A) 25\frac{2}{5}, B) 49\frac{4}{9}, C) 35\frac{3}{5}, D) 45\frac{4}{5}, E) 710\frac{7}{10}.

To crack this, we're going to need to roll up our sleeves and use a mix of trigonometric identities and geometric theorems. Think Pythagorean theorem, sine and cosine relationships, and maybe even some similar triangles. Don't worry, we'll break it down step by step. Geometry problems like these can seem daunting at first, but once you untangle them, they're super satisfying to solve!

Setting Up the Geometric Framework

Okay, let’s sketch this out (or imagine it vividly!). We have triangle ABC, and inside it, AD drops down to meet BC at a right angle. This immediately gives us two smaller right-angled triangles: ABD and ADC. These are goldmines for applying trigonometric ratios. Remember SOH-CAH-TOA? It's our best friend here.

We know AC is 10 cm and BD is 4 cm. The angle relationships are crucial: 2m(BAD)=m(ACB)=2α2m(BAD) = m(ACB) = 2\alpha. Let's denote the angle BAD as α\alpha. This makes angle ACB equal to 2α2\alpha. This setup is like a perfectly arranged puzzle – all the pieces are there, we just need to fit them together.

From the right triangle ABD, we can relate the sides using trigonometric functions of angle α\alpha. Similarly, in triangle ADC, we can relate sides using trigonometric functions, but this time of angle 2α2\alpha. The double angle is a little hint that we might need some double-angle formulas down the road. Think sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha = 2 \sin \alpha \cos \alpha – keep that in the back of your mind!

Applying Trigonometric Ratios

Let’s focus on triangle ABD first. We know BD is 4 cm. Let’s call AD 'h' (for height, makes sense, right?). So, in triangle ABD:

  • tan⁡α=BDAD=4h\tan \alpha = \frac{BD}{AD} = \frac{4}{h}

This gives us a neat relationship between α\alpha and h. Keep this equation handy! Now, let's swing over to triangle ADC. Here, we know AC is 10 cm, AD is 'h', and angle ACD is 2α2\alpha. We can use the sine and cosine functions here:

  • sin⁡2α=ADAC=h10\sin 2\alpha = \frac{AD}{AC} = \frac{h}{10}
  • cos⁡2α=DCAC=DC10\cos 2\alpha = \frac{DC}{AC} = \frac{DC}{10}

Okay, now we have expressions involving sin⁡2α\sin 2\alpha and cos⁡2α\cos 2\alpha. But wait, we need to find a way to link these back to α\alpha and our initial equation. This is where those double-angle formulas come into play. It’s like connecting the dots in a treasure map!

Leveraging Double-Angle Formulas

Remember the double-angle formula for sine? sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha = 2 \sin \alpha \cos \alpha. We also have the Pythagorean identity: sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1. These are like our secret weapons! We can rewrite sin⁡2α\sin 2\alpha as h10\frac{h}{10}, so now we have:

  • h10=2sin⁡αcos⁡α\frac{h}{10} = 2 \sin \alpha \cos \alpha

We need to express sin⁡α\sin \alpha and cos⁡α\cos \alpha in terms of what we know. Back in triangle ABD:

  • sin⁡α=BDAB=4AB\sin \alpha = \frac{BD}{AB} = \frac{4}{AB}
  • cos⁡α=ADAB=hAB\cos \alpha = \frac{AD}{AB} = \frac{h}{AB}

Aha! We're getting closer. Now, substitute these into our double-angle formula equation:

  • h10=2⋅4AB⋅hAB\frac{h}{10} = 2 \cdot \frac{4}{AB} \cdot \frac{h}{AB}
  • h10=8hAB2\frac{h}{10} = \frac{8h}{AB^2}

Notice that 'h' appears on both sides. Assuming h isn't zero (which it can’t be in a triangle!), we can cancel it out:

  • 110=8AB2\frac{1}{10} = \frac{8}{AB^2}
  • AB2=80AB^2 = 80
  • AB=80=45AB = \sqrt{80} = 4\sqrt{5}

Okay, we’ve found AB! This is a major breakthrough. Now we can plug this back into our sin⁡α\sin \alpha and cos⁡α\cos \alpha equations.

Solving for AD (h)

Let’s circle back to triangle ABD and use the Pythagorean theorem. We have AB and BD, and we're looking for AD (h):

  • AB2=AD2+BD2AB^2 = AD^2 + BD^2
  • (45)2=h2+42(4\sqrt{5})^2 = h^2 + 4^2
  • 80=h2+1680 = h^2 + 16
  • h2=64h^2 = 64
  • h=8h = 8

Boom! We've found h! AD is 8 cm. This is fantastic because now we can directly compute sin⁡2α\sin 2\alpha.

Calculating sin2α

Remember that we found sin⁡2α=h10\sin 2\alpha = \frac{h}{10}? We now know h is 8 cm, so:

  • sin⁡2α=810=45\sin 2\alpha = \frac{8}{10} = \frac{4}{5}

And there we have it! The value of sin⁡2α\sin 2\alpha is 45\frac{4}{5}. That matches option D! We nailed it!

Final Answer

So, guys, we solved it! The correct answer is D) 45\frac{4}{5}. We tackled this problem by breaking it down into smaller parts, using trigonometric ratios, double-angle formulas, and the Pythagorean theorem. Geometry problems like these are puzzles, and with a bit of patience and the right tools, they're totally solvable. Keep practicing, and you'll become a geometry whiz in no time!