Symmetric Triangle Angles: Find The Largest One

Hey math whizzes! Let's dive into a cool geometry problem today involving a triangle that's got some serious symmetry. We're dealing with a triangle called MNP, and we know the coordinates for two of its vertices: N is at (2, 2), and P is at (0, -2). The key piece of information here is that this triangle is symmetric about the y-axis. Our mission, should we choose to accept it, is to figure out the approximate measure of the largest angle in this triangle. So, grab your protractors (or just your awesome brainpower), and let's break this down!

Unpacking the Symmetry and Coordinates

So, we've got triangle MNP, and it's sitting pretty on the coordinate plane. We know N is at (2, 2) and P is at (0, -2). The big hint is that the triangle is symmetric about the y-axis. What does that really mean for us, guys? Symmetry about the y-axis means that if you were to fold the triangle along the y-axis, the two halves would match up perfectly. For a point (x, y), its reflection across the y-axis is (-x, y). Since our triangle MNP is symmetric about the y-axis, this tells us something crucial about the vertex M. If N is at (2, 2), its symmetric counterpart across the y-axis would be (-2, 2). Therefore, the vertex M must be at (-2, 2). Now we've got all three vertices: M(-2, 2), N(2, 2), and P(0, -2). This is awesome because now we can start calculating side lengths and angles!

Let's get calculating the lengths of the sides of our triangle MNP. We can use the distance formula, which is $d = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}$.

• Side MN: The distance between M(-2, 2) and N(2, 2). $MN = \sqrt{((2 - (-2))^2 + (2 - 2)^2)} = \sqrt{((2 + 2)^2 + 0^2)} = \sqrt{4^2} = \sqrt{16} = 4$. So, side MN has a length of 4 units.

• Side NP: The distance between N(2, 2) and P(0, -2). $NP = \sqrt{((0 - 2)^2 + (-2 - 2)^2)} = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$. We can simplify $\sqrt{20}$ to $2\sqrt{5}$. So, side NP has a length of $2\sqrt{5}$ units.

• Side MP: The distance between M(-2, 2) and P(0, -2). $MP = \sqrt{((0 - (-2))^2 + (-2 - 2)^2)} = \sqrt{((0 + 2)^2 + (-4)^2)} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$. So, side MP also has a length of $2\sqrt{5}$ units.

Okay, so we've found that MN = 4, NP = $2\sqrt{5}$, and MP = $2\sqrt{5}$. This is super interesting, guys! Since two sides of the triangle (NP and MP) have the same length, triangle MNP is an isosceles triangle. This makes sense given the symmetry we were told about. The y-axis is the perpendicular bisector of the base MN, and P lies on the y-axis, which is the axis of symmetry.

Now that we have the side lengths, we can determine which angle is the largest. In any triangle, the largest angle is opposite the longest side. Here, the sides are 4, $2\sqrt{5}$, and $2\sqrt{5}$. Let's compare 4 and $2\sqrt{5}$. Squaring both numbers, we get $4^2 = 16$ and $(2\sqrt{5})^2 = 4 \times 5 = 20$. Since 20 is greater than 16, $2\sqrt{5}$ is greater than 4. Therefore, the sides NP and MP are the longest sides. The angle opposite the longest sides will be the largest angles. The angle opposite side NP is $\angle M$, and the angle opposite side MP is $\angle N$. Since NP = MP, we know that $\angle M = \angle N$. The side MN is the shortest side, and it's opposite $\angle P$. So, $\angle P$ is likely to be the smallest angle.

Our goal is to find the largest angle, which means we need to find either $\angle M$ or $\angle N$. Since they are equal, let's focus on finding $\angle N$. We can use the Law of Cosines to find the angles. The Law of Cosines states that for a triangle with sides a, b, and c, and angle C opposite side c: $c^2 = a^2 + b^2 - 2ab \cos(C)$.

Let's find $\angle N$. The side opposite $\angle N$ is MP. So, $MP^2 = MN^2 + NP^2 - 2(MN)(NP)\cos(\angle N)$.

We have $MP = \sqrt{20}$, $MN = 4$, and $NP = \sqrt{20}$. Plugging these values in:

$(\sqrt{20})^2 = 4^2 + (\sqrt{20})^2 - 2(4)(\sqrt{20})\cos(\angle N)$

$20 = 16 + 20 - 8\sqrt{20}\cos(\angle N)$

$20 = 36 - 8\sqrt{20}\cos(\angle N)$

Now, let's rearrange to solve for $\cos(\angle N)$:

$8\sqrt{20}\cos(\angle N) = 36 - 20$

$8\sqrt{20}\cos(\angle N) = 16$

$\cos(\angle N) = \frac{16}{8\sqrt{20}} = \frac{2}{\sqrt{20}} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$

To find the angle $\angle N$, we need to take the inverse cosine (arccos) of $\frac{1}{\sqrt{5}}$.

$\angle N = \arccos(\frac{1}{\sqrt{5}})$

Using a calculator, $\frac{1}{\sqrt{5}} \approx \frac{1}{2.236} \approx 0.4472$.

$\angle N \approx \arccos(0.4472) \approx 63.43^{\circ}$

Since $\angle M = \angle N$, $\angle M \approx 63.43^{\circ}$.

Now, let's think about $\angle P$. We know that the sum of angles in a triangle is $180^{\circ}$. So, $\angle M + \angle N + \angle P = 180^{\circ}$.

Approximately, $63.43^{\circ} + 63.43^{\circ} + \angle P = 180^{\circ}$.

$126.86^{\circ} + \angle P = 180^{\circ}$.

$\angle P = 180^{\circ} - 126.86^{\circ} \approx 53.14^{\circ}$.

So, the angles are approximately $63.43^{\circ}$, $63.43^{\circ}$, and $53.14^{\circ}$.

Comparing these angles, the largest angle is approximately $63.43^{\circ}$. This corresponds to option C. It's pretty neat how the symmetry helps us identify the vertices and then calculate the angles.

Calculating Angles Using Slopes (An Alternative Approach)

Another way we can tackle finding the angles is by using the slopes of the lines forming the sides of the triangle. This can be a bit more intuitive for some folks, especially when dealing with coordinates. Remember, the slope ($m$) of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

We have our vertices: M(-2, 2), N(2, 2), and P(0, -2).

• Slope of MN ($m_{MN}$): $m_{MN} = \frac{2 - 2}{2 - (-2)} = \frac{0}{4} = 0$. This tells us that MN is a horizontal line, which aligns with our side length calculation where the y-coordinates were the same.

• Slope of NP ($m_{NP}$): $m_{NP} = \frac{-2 - 2}{0 - 2} = \frac{-4}{-2} = 2$.

• Slope of MP ($m_{MP}$): $m_{MP} = \frac{-2 - 2}{0 - (-2)} = \frac{-4}{2} = -2$.

Now, how do we use these slopes to find angles? We can use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$: $\tan(\theta) = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. This formula gives us the acute angle between the lines. If we need the obtuse angle, we can subtract it from 180 degrees.

Let's find the angle at vertex P, $\angle P$. This is the angle between lines NP and MP. So, $m_1 = m_{NP} = 2$ and $m_2 = m_{MP} = -2$.

$\tan(\angle P) = |\frac{2 - (-2)}{1 + (2)(-2)}| = |\frac{2 + 2}{1 - 4}| = |\frac{4}{-3}| = \frac{4}{3}$.

$\angle P = \arctan(\frac{4}{3}) \approx 53.13^{\circ}$.

This matches what we found using the Law of Cosines, which is awesome!

Now, let's find the angle at vertex N, $\angle N$. This is the angle between lines MN and NP. So, $m_1 = m_{MN} = 0$ and $m_2 = m_{NP} = 2$.

$\tan(\angle N) = |\frac{0 - 2}{1 + (0)(2)}| = |\frac{-2}{1}| = |-2| = 2$.

$\angle N = \arctan(2) \approx 63.43^{\circ}$.

This also matches our previous calculation for $\angle N$! Since the triangle is isosceles with MP = NP, we know $\angle M = \angle N$. So $\angle M \approx 63.43^{\circ}$.

So, the angles are approximately $63.43^{\circ}$, $63.43^{\circ}$, and $53.13^{\circ}$. The largest angle is indeed approximately $63.43^{\circ}$.

Identifying the Largest Angle and Final Answer

We've calculated the angles of triangle MNP using two different methods, and both give us consistent results. The angles are approximately $63.4^{\circ}$, $63.4^{\circ}$, and $53.1^{\circ}$.

When we compare these values, it's clear that the largest angle is approximately $63.4^{\circ}$. This value is closest to option C, which is $63.6^{\circ}$. The slight difference is likely due to rounding in our calculations or the precision of the given options. However, $63.4^{\circ}$ is definitely closest to $63.6^{\circ}$.

To recap, the symmetry about the y-axis was the key to finding the coordinates of vertex M. Once we had all three vertices, we could determine the lengths of the sides and identify that it's an isosceles triangle. Using either the Law of Cosines or the slopes of the sides, we calculated the angles. We found that the two base angles (at M and N) are larger than the vertex angle (at P). The largest angle in the triangle MNP is approximately $63.4^{\circ}$, which makes option C the correct answer. Great job working through this problem, guys! Keep practicing, and you'll be a geometry guru in no time!