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Suppose a triangle has two sides of length 2 and 3, and the angle between these two sides is [tex][tex]$60^{\circ}$[/tex][/tex]. What is the length of the third side of the triangle?

A. [tex][tex]$\sqrt{3}$[/tex][/tex]
B. [tex][tex]$2 \sqrt{3}$[/tex][/tex]
C. [tex][tex]$\sqrt{7}$[/tex][/tex]
D. [tex]2[/tex]


Sagot :

Let's solve for the length of the third side of a triangle given two sides and the included angle.

For a triangle with sides of lengths 2 and 3, and an included angle of [tex]\(60^\circ\)[/tex], we will use the Law of Cosines to determine the length of the third side.

The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(C) \][/tex]

where:
- [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the lengths of the two given sides,
- [tex]\( C \)[/tex] is the included angle, and
- [tex]\( c \)[/tex] is the length of the third side.

Plugging in the given values:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( C = 60^\circ \)[/tex]

First, we convert the angle from degrees to radians because the cosine function typically requires the input angle to be in radians:
[tex]\[ 60^\circ = \frac{\pi}{3} \text{ radians} \][/tex]
[tex]\[ \cos(60^\circ) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \][/tex]

Now, substitute these values into the Law of Cosines formula:
[tex]\[ c^2 = 2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \cos\left(\frac{\pi}{3}\right) \][/tex]

Calculate each part step-by-step:
[tex]\[ 2^2 = 4 \][/tex]
[tex]\[ 3^2 = 9 \][/tex]
[tex]\[ 2 \cdot 2 \cdot 3 \cdot \frac{1}{2} = 6 \][/tex]

Now, substitute these into the equation:
[tex]\[ c^2 = 4 + 9 - 6 \][/tex]
[tex]\[ c^2 = 13 - 6 \][/tex]
[tex]\[ c^2 = 7 \][/tex]

To find [tex]\( c \)[/tex], take the square root of both sides:
[tex]\[ c = \sqrt{7} \][/tex]

Therefore, the length of the third side is [tex]\( \sqrt{7} \)[/tex]. Thus, the correct answer is:

C. [tex]\( \sqrt{7} \)[/tex]