Two corners of a triangle have angles of #pi / 3 # and # pi / 2 #. If one side of the triangle has a length of #7 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #pi  /  3  # and #  pi   /  2  #.  If one side of the triangle has a length of #7  #, what is the longest possible perimeter of the triangle?

Sophia Alfred · Accepted Answer

Longest possible perimeter is #33.124#.

As two angles are #pi/2# and #pi/3#, the third angle is #pi-pi/2-pi/3=pi/6#. This is the least angle and hence side opposite this is smallest. As we have to find longest possible perimeter, whose one side is #7#, this side must be opposite the smallest angle i.e. #pi/6#. Let other two sides be #a# and #b#. Hence using sine formula #7/sin(pi/6)=a/sin(pi/2)=b/sin(pi/3)# or #7/(1/2)=a/1=b/(sqrt3/2)# or #14=a=2b/sqrt3# Hence #a=14# and #b=14xxsqrt3/2=7xx1.732=12.124# Hence, longest possible perimeter is #7+14+12.124=33.124#

Sadie Allen · Answer

undefined To find the longest possible perimeter of the triangle, we first need to determine the lengths of the other two sides. We can use the Law of Sines to find these lengths. The Law of Sines states that for any triangle with sides of lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ respectively, the following ratio holds:

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

Given that two angles of the triangle are $ \frac{\pi}{3} $ and $ \frac{\pi}{2} $, we can find the third angle using the fact that the sum of angles in a triangle is $ \pi $. Therefore, the third angle is $ \pi - \frac{\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6} $.

Now, let's use the Law of Sines with the known side length of 7 and the corresponding angles:

$$\frac{\sin(\frac{\pi}{3})}{7} = \frac{\sin(\frac{\pi}{6})}{x}$$

Solving for $x$, we find:

$$x = \frac{7 \cdot \sin(\frac{\pi}{6})}{\sin(\frac{\pi}{3})} = \frac{7 \cdot \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{7}{\sqrt{3}}$$

Thus, the length of the second side is $ \frac{7}{\sqrt{3}} $.

Using the Law of Sines again to find the length of the third side:

$$\frac{\sin(\frac{\pi}{2})}{7} = \frac{\sin(\frac{\pi}{6})}{y}$$

Solving for $y$, we find:

$$y = \frac{7 \cdot \sin(\frac{\pi}{6})}{\sin(\frac{\pi}{2})} = \frac{7 \cdot \frac{1}{2}}{1} = \frac{7}{2}$$

Now that we have all three side lengths, we can calculate the perimeter of the triangle:

$$P = 7 + \frac{7}{\sqrt{3}} + \frac{7}{2}$$

$$P = 7 + \frac{14}{\sqrt{3}}$$

This is the longest possible perimeter of the triangle.

Theodore Abad · Answer

undefined To find the longest possible perimeter of the triangle, we need to consider the sum of the lengths of the other two sides. Since we know one side is 7 units long, we can use trigonometric ratios to find the lengths of the other two sides.

Let's label the triangle as ABC, where angle A = π/3, angle B = π/2, and side AB = 7 units.

Using the sine and cosine ratios:

- For angle A (π/3):
  - sin(π/3) = opposite / hypotenuse
  - sin(π/3) = BC / 7
  - BC = 7 * sin(π/3)

- For angle B (π/2):
  - cos(π/2) = adjacent / hypotenuse
  - cos(π/2) = AC / 7
  - AC = 7 * cos(π/2)

Now, calculate the lengths of sides BC and AC:

- BC = 7 * sin(π/3) = 7 * √3 / 2
- AC = 7 * cos(π/2) = 0

Since AC is 0, it means the triangle is degenerate (i.e., a line segment), and it's not a valid triangle.

Thus, the longest possible perimeter of the triangle is the sum of AB and BC:
Perimeter = AB + BC = 7 + 7 * √3 / 2 = 7 + 7 * √3 / 2 = 7 * (1 + √3 / 2) ≈ 14.55 units