Problem 22 - Algebra 2 and Trigonometry

Question

In \(11-22,\) solve each triangle, that is, find the measures of the remaining three parts of the triangle to the nearest integer or the nearest degree. In \(\triangle A B C, a=15, b=25,\) and \(c=12\)

Step-by-Step Solution

Answer

\( A \approx 25^\circ, B \approx 45^\circ, C \approx 110^\circ \)

Step 1: Calculate the semi-perimeter

To apply Heron's formula later, we first need the semi-perimeter of the triangle. The semi-perimeter \( s \) is half of the perimeter. Calculate as follows: \[ s = \frac{a+b+c}{2} = \frac{15+25+12}{2} = 26 \]

Step 2: Find the area using Heron's Formula

Now, using Heron's formula, we can find the area \( A \) of the triangle:\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]Substitute the values:\[ A = \sqrt{26(26-15)(26-25)(26-12)} \]\[ A = \sqrt{26 \times 11 \times 1 \times 14} = \sqrt{4004} \approx 63.32 \]

Step 3: Applying Sine Rule to find an angle

Since it's difficult to directly use the cosine rule due to possible rounding issues earlier, use the Sine rule to find one of the angles.\[ \frac{a}{\sin A} = \frac{2 \times A}{a \times b} \]For simplicity, find angle \( A \):Using area again in \( \sin A = \frac{A}{\frac{1}{2} \times b \times c} \):\[ \sin A = \frac{63.32}{\frac{1}{2} \times 25 \times 12} = 0.4213 \]Therefore, \( A \approx \arcsin(0.4213) \approx 25^\circ \).

Step 4: Apply the Cosine Rule to find another angle

Using the cosine rule to find angle \( B \):\[ b^2 = a^2 + c^2 - 2ac \cdot \cos B \]\[ 25^2 = 15^2 + 12^2 - 2 \cdot 15 \cdot 12 \cdot \cos B \]\[ 625 = 225 + 144 - 360 \cdot \cos B \]\[ 625 = 369 - 360 \cdot \cos B \]\[ \cos B = \frac{369 - 625}{-360} \approx 0.7111 \]\( B \approx \arccos(0.7111) \approx 45^\circ \).

Step 5: Determine the last angle C

The sum of angles in a triangle is always \( 180^\circ \). Hence, we can find angle \( C \) by:\[ C = 180^\circ - A - B \]\[ C = 180^\circ - 25^\circ - 45^\circ \approx 110^\circ \].

Key Concepts

Heron's Formula

Heron's formula is a powerful tool for finding the area of a triangle when you know the lengths of all three sides. Unlike the basic area formula for triangles, which requires the base and height, Heron's formula gets the job done with just the side lengths. This can be super handy!
To use Heron's formula, first find the semi-perimeter of the triangle, which is simply half the perimeter. The formula is:

• The semi-perimeter, denoted as \( s \), is calculated as \( s = \frac{a+b+c}{2} \).

After calculating the semi-perimeter, plug it into Heron's formula:

• The area \( A \) of the triangle is calculated as \( A = \sqrt{s(s-a)(s-b)(s-c)} \).

This formula may look a bit complex, but with some practice, you'll find it's a straightforward way to determine an unknown triangle area.

Sine Rule

The Sine Rule (or Law of Sines) is useful for solving triangles, especially when dealing with oblique triangles. This rule relates the sides of a triangle to the sines of its angles. Basically, for any triangle that has a side opposite a known angle, the Sine Rule can help you find missing angles or sides.
The rule is given by:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

Here's how it works: if you know two angles and one side of a triangle, or two sides and a non-included angle, the Sine Rule is your friend. It's incredibly useful for finding unknown angles, especially after using Heron’s formula to find the area, which then helps calculate a specific angle using its sine. Make sure to use it when calculating non-right triangles, especially when some angles are easier to find than others.

Cosine Rule

The Cosine Rule (Or Law of Cosines) is another essential rule for solving triangles, particularly when the triangle is not right-angled. It is best used when you know either all three sides or two sides and the included angle. It's like the big brother of the Pythagorean theorem but works for all types of triangles, not just right-angled ones.
The Cosine Rule can be expressed as follows:

• \( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \)
• \( a^2 = b^2 + c^2 - 2bc \cdot \cos(A) \)
• \( b^2 = a^2 + c^2 - 2ac \cdot \cos(B) \)

This rule allows you to find unknown sides or angles by re-arranging the formula based on what's known. It's particularly helpful when calculating the remaining angles of a triangle after the area has been found and one angle is known.

Semiperimeter

The semiperimeter is a crucial concept to understand when using Heron's formula for calculating the area of a triangle. It sounds a bit fancy but is actually quite simple. The semi-perimeter is just half the perimeter of a triangle.
Here’s how you find it:

• Calculate the perimeter of the triangle by adding up all three side lengths: \( a + b + c \).
• Divide this sum by 2 to get the semi-perimeter: \( s = \frac{a+b+c}{2} \).

The semi-perimeter is a stepping stone to finding the area with Heron’s formula. Think of it as preparing the triangle for a deeper calculation! Without it, Heron's formula wouldn't work, as the semi-perimeter is an integral part of the formula. It's one of those small steps that have a big impact when working through triangle problems.