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Chapter 4: Problem 36

Find the perimeter of an isosceles triangle that has two sides of length 8 and a \(130^{\circ}\) angle between those two sides.

Short Answer

Expert verified
The perimeter of the given isosceles triangle is approximately 31.98.

Step by step solution

01

Understand the Problem and Given Information

We are given an isosceles triangle with two equal sides of length 8 and a 130-degree angle between those two sides. Let's denote the vertices of the triangle as A, B, and C, where A and C are the two equal sides, and B is the angle between them.
02

Apply the Law of Cosines

We can apply the Law of Cosines to find the length of the base (side BC). The Law of Cosines states that for any triangle with sides a, b, and c, and opposite angles A, B, and C respectively, the following relationship holds: \[c^2 = a^2 + b^2 - 2ab\cos C\] In our case: - Side AB = Side AC = 8 (isosceles triangle's equal sides) - Angle B = 130 degrees We want to find Side BC. Let's denote the length of Side BC as x. So, we will use the following equation: \[x^2 = 8^2 + 8^2 - 2(8)(8)\cos 130^\circ\]
03

Calculate the Length of Side BC

First, we need to convert 130 degrees to radians, as the cosine function in calculators uses radians. We can use the following conversion formula: Radians = Degrees 脳 \(\frac{\pi}{180}\) Convert 130 degrees to radians: Radians = \(130\times\frac{\pi}{180} = \frac{13\pi}{9}\) Now, we can substitute the values into the equation: \[x^2 = 64 + 64 - (128)(\cos\frac{13\pi}{9})\] Calculate the value of the cosine: \(\cos\frac{13\pi}{9} \approx -0.6428\) Now, plug this value back into the equation for x: \[x^2 = 128 - 128(-0.6428)\] \[x^2 \approx 255.5\] Now, we take the square root of 255.5 to find the length of side BC: \[x = \sqrt{255.5} \approx 15.98\] So, the length of side BC is approximately 15.98.
04

Calculate the Perimeter

Finally, we calculate the perimeter by adding all the side lengths: Perimeter = Side AB + Side AC + Side BC Perimeter = \(8 + 8 + 15.98 \approx 31.98\) So, the perimeter of the given isosceles triangle is approximately 31.98.

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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.] \(\sin \frac{17 \pi}{8}\)

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.] \(\tan \left(-\frac{3 \pi}{8}\right)\)

A surveyor wishes to measure the distance between points \(A\) and \(B\), but buildings between \(A\) and \(B\) prevent a direct measurement. Thus the surveyor moves 50 meters perpendicular to the line \(A B\) to the point \(C\) and measures that angle \(B C A\) is \(87^{\circ} .\) What is the distance between the points \(A\) and \(B ?\)

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.] \(\sin \frac{\pi}{12}\)

Assume the surface of the earth is a sphere with diameter 7926 miles. Approximately how far does a ship travel when sailing along the equator in the Pacific Ocean from longitude \(170^{\circ}\) west to longitude \(120^{\circ}\) west?
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