Solving Triangles: Perimeter & Area Calculations

Hey guys! Today, we're diving into the exciting world of triangles! Specifically, we're going to tackle two problems where we need to solve triangles, calculate their perimeters, and find their areas. Sounds like fun, right? Don't worry, we'll break it down step-by-step so it's super easy to follow. Let's get started!

Problem 1: Triangle ABC with a = 88 m, Â = 28°40', and Ĉ = 40°28'

In this first problem, we're given one side (a = 88 meters) and two angles (Â = 28°40' and Ĉ = 40°28') of triangle ABC. Our mission, should we choose to accept it (and we do!), is to find the remaining sides and angles, and then calculate the perimeter and area of the triangle. So, how do we even start? Well, the first thing we need to do is find the remaining angle.

Finding the Missing Angle

The key here is to remember that the sum of the angles in any triangle is always 180 degrees. This is a fundamental concept in geometry, and it's going to be our best friend in solving this problem. We know two angles already, so finding the third one is just a simple subtraction problem.

Let's call the missing angle angle B. We can set up the equation like this:

 + Ĉ + B = 180°

Now, let's plug in the values we know:

28°40' + 40°28' + B = 180°

Adding the two angles, we get:

69°08' + B = 180°

To isolate B, we subtract 69°08' from both sides:

B = 180° - 69°08'

This gives us:

B = 110°52'

Awesome! We've found our missing angle. Now we know all three angles of the triangle.

Finding the Missing Sides: The Law of Sines

Alright, we've got all the angles, but we still need to find the lengths of the sides b and c. To do this, we're going to use a nifty little tool called the Law of Sines. The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is constant for all sides and angles in a triangle. In mathematical terms, it looks like this:

a / sin(Â) = b / sin(B) = c / sin(Ĉ)

This might look a little intimidating at first, but don't worry, it's actually quite straightforward to use. We already know side 'a' and all the angles, so we can use the Law of Sines to find sides 'b' and 'c'.

Let's start with side 'b'. We can set up the following proportion:

88 m / sin(28°40') = b / sin(110°52')

To solve for 'b', we can cross-multiply:

b * sin(28°40') = 88 m * sin(110°52')

Now, divide both sides by sin(28°40'):

b = (88 m * sin(110°52')) / sin(28°40')

Using a calculator, we find:

b ≈ (88 m * 0.9344) / 0.4804

b ≈ 171.06 m

Great! We've found side 'b'. Now, let's find side 'c' using the same approach:

88 m / sin(28°40') = c / sin(40°28')

Cross-multiply:

c * sin(28°40') = 88 m * sin(40°28')

Divide both sides by sin(28°40'):

c = (88 m * sin(40°28')) / sin(28°40')

Using a calculator:

c ≈ (88 m * 0.6479) / 0.4804

c ≈ 118.68 m

Fantastic! We've now found all the sides and all the angles of triangle ABC. Pat yourself on the back – that's a big accomplishment!

Calculating the Perimeter

The perimeter of a triangle is simply the sum of the lengths of its sides. Easy peasy! We know the lengths of all three sides now, so we can just add them up:

Perimeter = a + b + c

Perimeter = 88 m + 171.06 m + 118.68 m

Perimeter ≈ 377.74 m

So, the perimeter of triangle ABC is approximately 377.74 meters.

Calculating the Area: Heron's Formula

Now, for the grand finale: calculating the area of the triangle. We have a couple of options here, but since we know all three sides, we're going to use Heron's Formula. Heron's Formula is a super useful tool for finding the area of a triangle when you know the lengths of all three sides. It looks like this:

Area = √(s * (s - a) * (s - b) * (s - c))

Where 's' is the semi-perimeter of the triangle, which is half of the perimeter:

s = Perimeter / 2

First, let's calculate the semi-perimeter:

s ≈ 377.74 m / 2

s ≈ 188.87 m

Now we can plug everything into Heron's Formula:

Area ≈ √(188.87 m * (188.87 m - 88 m) * (188.87 m - 171.06 m) * (188.87 m - 118.68 m))

Area ≈ √(188.87 m * 100.87 m * 17.81 m * 70.19 m)

Area ≈ √(239,246,544.96 m^4)

Area ≈ 15,467.60 m²

Therefore, the area of triangle ABC is approximately 15,467.60 square meters. Wow! We did it! We successfully solved the first triangle and calculated its perimeter and area.

Problem 2: Triangle ABC with a = 340 cm, b = 50 cm, and Ĉ = 46°20'

Okay, guys, let's move on to the second problem! This time, we're given two sides (a = 340 cm and b = 50 cm) and one angle (Ĉ = 46°20'). Again, we need to find the remaining sides and angles, and then calculate the perimeter and area. Are you ready for another round?

Finding the Missing Sides: The Law of Cosines

In this case, since we have two sides and the included angle (the angle between them), we're going to use the Law of Cosines to find the missing side 'c'. The Law of Cosines is a generalization of the Pythagorean theorem and is super useful when you don't have a right triangle. It looks like this:

c² = a² + b² - 2ab * cos(Ĉ)

Let's plug in the values we know:

c² = 340² cm² + 50² cm² - 2 * 340 cm * 50 cm * cos(46°20')

c² = 115600 cm² + 2500 cm² - 34000 cm² * cos(46°20')

Using a calculator, we find:

c² = 118100 cm² - 34000 cm² * 0.6905

c² = 118100 cm² - 23477 cm²

c² = 94623 cm²

Now, take the square root of both sides to find 'c':

c = √94623 cm²

c ≈ 307.61 cm

Excellent! We've found the missing side 'c'. Now we have all three sides of the triangle.

Finding the Missing Angles: The Law of Sines (Again!)

Now that we have all three sides, we can use the Law of Sines again to find the missing angles  and B. Let's start with angle Â:

a / sin(Â) = c / sin(Ĉ)

340 cm / sin(Â) = 307.61 cm / sin(46°20')

Cross-multiply:

340 cm * sin(46°20') = 307.61 cm * sin(Â)

Divide both sides by 307.61 cm:

sin(Â) = (340 cm * sin(46°20')) / 307.61 cm

sin(Â) = (340 cm * 0.7222) / 307.61 cm

sin(Â) ≈ 0.7984

To find angle Â, we need to take the inverse sine (arcsin) of 0.7984:

 = arcsin(0.7984)

 ≈ 53.00°

Great! We've found angle Â. Now, to find angle B, we can use the fact that the sum of the angles in a triangle is 180°:

 + B + Ĉ = 180°

53.00° + B + 46°20' = 180°

B = 180° - 53.00° - 46°20'

B = 80.67°

Fantastic! We've now found all the sides and angles of the second triangle. Time to calculate the perimeter and area!

Calculating the Perimeter

Just like before, the perimeter is the sum of the lengths of the sides:

Perimeter = a + b + c

Perimeter = 340 cm + 50 cm + 307.61 cm

Perimeter ≈ 697.61 cm

So, the perimeter of triangle ABC in the second problem is approximately 697.61 centimeters.

Calculating the Area: Heron's Formula (Again!)

Let's use Heron's Formula again to find the area. First, we need the semi-perimeter:

s = Perimeter / 2

s ≈ 697.61 cm / 2

s ≈ 348.81 cm

Now, plug everything into Heron's Formula:

Area = √(s * (s - a) * (s - b) * (s - c))

Area ≈ √(348.81 cm * (348.81 cm - 340 cm) * (348.81 cm - 50 cm) * (348.81 cm - 307.61 cm))

Area ≈ √(348.81 cm * 8.81 cm * 298.81 cm * 41.20 cm)

Area ≈ √(360,031,234.84 cm^4)

Area ≈ 18,974.49 cm²

Therefore, the area of triangle ABC in the second problem is approximately 18,974.49 square centimeters.

Conclusion

Wow, guys! We've successfully solved two challenging triangle problems, calculated their perimeters, and found their areas. We used the Law of Sines, the Law of Cosines, and Heron's Formula – some pretty powerful tools in the world of geometry! I hope you found this step-by-step guide helpful and that you're feeling more confident in your triangle-solving abilities. Keep practicing, and you'll be a triangle master in no time! Remember, geometry can be fun! Until next time, keep those angles acute and your sides in proportion!