# Solve each triangle using the labels as shown in Illustration 1 (round lengths of sides to three significant digits and angles to the nearest tenth of a degree): ILLUSTRATION 1 B = 18.5 ° , a = 1680 m , b = 1520 m

### Elementary Technical Mathematics

12th Edition
Dale Ewen
Publisher: Cengage Learning
ISBN: 9781337630580

Chapter
Section

### Elementary Technical Mathematics

12th Edition
Dale Ewen
Publisher: Cengage Learning
ISBN: 9781337630580
Chapter 14, Problem 15R
Textbook Problem
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## Solve each triangle using the labels as shown in Illustration 1 (round lengths of sides to three significant digits and angles to the nearest tenth of a degree):ILLUSTRATION 1 B = 18.5 ° , a = 1680  m , b = 1520  m

To determine

To calculate: Each angle and dimension of triangle using the labels as shown in the given figure and the provided data is B=18.5°,a=1680m,b=1520cm

### Explanation of Solution

Given Information:

The given information: B=18.5°,a=1680m,b=1520cm and the triangle is give below,

Formula used:

The sine rule for triangle is given below as,

asinA=bsinB=csinC

Calculation:

Consider the give information about the triangle,

Now find the dimensions required by using the law of sine,

As the length of b and angle of B are given,

Find the length of a:

bsinB=asinA1520sin18.5°=1680sinA°15200.3173=1680sinAsinA=0.3507A=20.5°

Similarly, find the length of c:

First find the angle of C, by subtracting the angles of A and B from 180°:

Angle C=180°18.5°20.5°=141°

Now, find the length of c:

csinC=bsinBcsin141°=1520sin18

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