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Chapter 4: Problem 41

Find a formula for the perimeter of an isosceles triangle that has two sides of length \(c\) with angle \(\theta\) between those two sides.

Short Answer

Expert verified
The formula for the perimeter of an isosceles triangle with two sides of length \(c\) and angle 胃 between those sides is: \[P = \sqrt{2c^2 - 2c^2 \cos(胃)} + 2c\]

Step by step solution

01

Understand the properties of an isosceles triangle

An isosceles triangle has two sides of equal length, which we'll call the congruent sides. In our problem, these sides have a length of \(c\). The angle between these congruent sides is 胃.
02

Find the length of the base side

To find the length of the base side, let's call it \(b\), we can use the Law of Cosines. The Law of Cosines states that, for any triangle with sides of length a, b, and c, and the angle opposite to side c is equal to 纬, the following relationship holds: \[c^2 = a^2 + b^2 - 2ab \cos(纬)\] In our case, the triangle sides are as follows: \(a=c, b, c\). The angle between the congruent sides, 胃, is opposite to the base side b. So, the Law of Cosines becomes: \[b^2 = c^2 + c^2 - 2c^2 \cos(胃)\]
03

Simplify the equation for the base side

Now let's simplify our equation: \[b^2 = 2c^2 - 2c^2 \cos(胃)\] To find the length of the base side b, take the square root of both sides: \[b = \sqrt{2c^2 - 2c^2 \cos(胃)}\]
04

Calculate the perimeter of the isosceles triangle

The perimeter of a triangle is the sum of the lengths of all its sides. In our case, the perimeter P, of the isosceles triangle can be represented as: \[P = b + 2c\] Substituting the equation for the base side b found in step 3, we have: \[P = \sqrt{2c^2 - 2c^2 \cos(胃)} + 2c\] This is the formula for the perimeter of an isosceles triangle with two sides of length \(c\) and angle 胃 between those sides.

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Most popular questions from this chapter

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\tan \left(-\frac{\pi}{8}\right)\)

Assume the surface of the earth is a sphere with radius 3963 miles. The latitude of \(a\) point \(P\) on the earth's surface is the angle between the line from the center of the earth to \(P\) and the line from the center of the earth to the point on the equator closest to \(P\), as shown below for latitude \(40^{\circ} .\) Cleveland has latitude \(41.5^{\circ}\) north. Find the radius of the circle formed by the points with the same latitude as Cleveland.

Explain why \(\sin 3^{\circ}+\sin 357^{\circ}=0\).

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In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\sin \frac{25 \pi}{12}\)
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