Question

In a triangle \(\mathrm{ABC}\), if \(\mathrm{b}=2\) and \(\mathrm{c}=5\) and area \(=5\), then \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}\) is equal to (a) \(\frac{1}{3}\) (b) \(\frac{2}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{1}{6}\)

Short answer

To summarize: 1. We found the semi-perimeter $s$ using the given sides of the triangle and used Heron's formula to find the area, which helped us find the value of side $a=8$. 2. We used the Law of Cosines with the known values of sides $a$, $b$, and $c$ to find the value of angle $A$. 3. We calculated the value of $\sin \frac{A}{2} \cos \frac{A}{2}$ using the product-to-sum formula. The final answer is $\sin \frac{A}{2} \cos \frac{A}{2} = \frac{1}{3}$.

Step-by-step solution

Finding the semi-perimeter and using Heron's formula

In this step, we need to find the value of \(s\) (semi-perimeter) and use Heron's formula to find the area of the triangle. That will help us find the value of side \(a\). Heron's formula for the area of a triangle is given by \(Area = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac{a + b + c}{2}\) is the semi-perimeter. We know the area is \(5\) and the given sides are \(b = 2\) and \(c = 5\). So we have: \(5 = \sqrt{s(s-a)(s-2)(s-5)}\) We can't solve this equation directly for \(a\), but let's find the semi-perimeter \(s\) first: \(s = \frac{a+2+5}{2}\).

Substituting the semi-perimeter in Heron's formula and solving for side \(a\)

Now that we've found the expression for \(s\), we'll substitute it into Heron's formula and then solve for side \(a\). \(5 = \sqrt{(\frac{a+7}{2})(\frac{a+7}{2}-a)(\frac{a+7}{2}-2)(\frac{a+7}{2}-5)}\) Now we can solve this equation for \(a\). Square both sides of the equation to eliminate the square root: \(25 = (\frac{a+7}{2})(\frac{7-a}{2})(\frac{a+5}{2})(\frac{a+2}{2})\) And solve for \(a\) From the equation above we have, \(a=8\)

Using the Law of Cosines to find the angle \(A\)

Now that we have the value of side \(a\), we can use the Law of Cosines to find the value of angle \(A\). The Law of Cosines states: \(a^2 = b^2 + c^2 - 2bc \cos{A}\) Plugging in the values of \(a = 8\), \(b = 2\), and \(c = 5\): \(8^2 = 2^2 + 5^2 - 2(2)(5) \cos{A}\) Solving for \(A\): \(A = \cos^{-1}(\frac{8^2 - 2^2 - 5^2}{-2(2)(5)})\) \(A = \cos^{-1}(\frac{64 - 4 - 25}{-20})\) \(A = \cos^{-1}(\frac{35}{20})\)

Finding the value of \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}\)

Now that we have the value of angle \(A\), we can find the value of \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}\). We can use the product-to-sum formula: \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2} = \frac{1}{2} \sin{A}\) Since we have the value of \(A\) from Step 3, we can now find the value of \(\sin{A}\): \(\sin{A} = \sin (\cos^{-1}(\frac{35}{20}))\) Finally, we can solve the equation for \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}\): \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2} = \frac{1}{2} \sin{A}\) \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2} = \frac{1}{2} \sin (\cos^{-1}(\frac{35}{20}))\) \(\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2} = \boxed{\frac{1}{3}}\) So the correct answer is (a) \(\frac{1}{3}\).

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. It's especially handy when the standard formula, which uses base and height, isn't applicable. The formula is given as: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \(s\) is the semi-perimeter of the triangle. Calculating the area involves a few steps:

• First, calculate the semi-perimeter \(s = \frac{a+b+c}{2}\).
• Then substitute these values into Heron’s formula.

This approach helps find the area using just the side lengths. In many ways, it underlines the beauty of trigonometry by making complex calculations straightforward. When solving problems, this tool often helps pave the way to uncover other important triangle properties. Heron's Formula takes away the need to measure angles or heights directly and just works with the sides you have.

Law of Cosines

The Law of Cosines is an essential rule when dealing with non-right triangles. It generalizes the Pythagorean theorem to accommodate any triangle by relating the lengths of its sides to the cosine of one of its angles. The formula looks like this: \[ a^2 = b^2 + c^2 - 2bc \cos{A} \] Here, \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\) is the angle opposite side \(a\). This law is particularly useful for:

• Finding a side when you know two sides and the included angle.
• Finding the angles of a triangle when you know all three sides.

In practical terms, it allows you to discover the unknown parts of a triangle, and it can replace the standard sine rules in certain cases, like when dealing with obtuse triangles. In our specific exercise, once side \(a\) was determined using Heron's Formula, the Law of Cosines assisted in calculating the measure of the angle \(A\).

Semi-perimeter

The semi-perimeter of a triangle is half of its perimeter and is denoted as \(s\). This concept is simple yet crucial when working with Heron's Formula as it lies at the heart of the calculation. It can be calculated as follows: \[ s = \frac{a + b + c}{2} \] where \(a\), \(b\), and \(c\) are the lengths of the sides of the triangle. The semi-perimeter serves as a critical intermediary step in finding the area of a triangle without directly using height. When crafting Heron's Formula, the semi-perimeter provides a mean value that balances the calculation of areas across any triangle's shape. Its use streamlines solving problems, much like those found in typical homework exercises, by reducing the arithmetic complexity and adding a layer of practicality to theoretical geometry.