You will do this by using the Law of Cosines to find the largest angle, then using the Law of Sines to find the middle angand then subtract to find the smallest angle. (There are other ways to proceed - you might say they are better ways!However, I am insisting that you do it in this manner.)I strongly suggest you look up the Law of Sines and the Law of Cosines in case you've forgotten them. Also, don't forgetabout convert ing from radians to degrees. In addition, you should Google the trigonometric and inverse trigonometricfunctions that you will need for your calculations.When you run your program, a sample run might look like this (where the user inputs 5, 4 and 3):Enter largest side length: 5Enter middle side length: 4Enter smallest side length: 3The angles are:90.053.1301023541559836.86989764584401or this (where the user inputs 4.1, 2.6 and 2.4):Enter largest side length: 4.1Enter middle side length: 2.6Enter smallest side leng th: 2.4The angles are:110. 10550978791636.548442826074876331346047386009 13Hints: the largest angle is opposite the largest side, the middle angle is opposite the middle side, and the smallest angle isopposite the smallest side. Also, be very careful about order of operations!Specifications: your program must• ask the user to enter three side lengths in descending order (two consecutive equal sides is valid also), each of which canbe any positive numberorder, or negative numbers, then your program does not need to work. You may also assume that the user entersnumbers that are the sidelengths of a real triangle! That is, you don't have to worry about the user entering 100, 2 and1, because no real triangle has those three sidelengths (remember the triangle inequality?).even one with decimals. You may assume that the user obeys if the user enters sides out of• correctly calculate the three angles in the triangle, in degrees, under the assumptions ment ioned above.• use the strategy I outlined above: Law of Cosines to find the largest angle, Law of Sines to find the middle angle, andsubtracting to find the smallest angle.• print out the three angles in degrees, each on a separate line. You don't have to worry about how many decimalplaces they are out put with.that is, if they enter side lengths that are either not all positive, aren't entered inChallenge: if the user is disobedientdescending order, or don't satisfy the triangle inequality - make the program print out a niessage saying so instead of tryingto perform the computations (which could result in an error). You might need to read ahead to do this.3MacBook Air1)F7F9F10F11F12

Answered: When you run your program, a sample run…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Question

You will do this by using the Law of Cosines to find the largest angle, then using the Law of Sines to find the middle ang
and then subtract to find the smallest angle. (There are other ways to proceed - you might say they are better ways!
However, I am insisting that you do it in this manner.)
I strongly suggest you look up the Law of Sines and the Law of Cosines in case you've forgotten them. Also, don't forget
about convert ing from radians to degrees. In addition, you should Google the trigonometric and inverse trigonometric
functions that you will need for your calculations.
When you run your program, a sample run might look like this (where the user inputs 5, 4 and 3):
Enter largest side length: 5
Enter middle side length: 4
Enter smallest side length: 3
The angles are:
90.0
53.13010235415598
36.86989764584401
or this (where the user inputs 4.1, 2.6 and 2.4):
Enter largest side length: 4.1
Enter middle side length: 2.6
Enter smallest side leng th: 2.4
The angles are:
110. 105509787916
36.548442826074876
331346047386009 13
Hints: the largest angle is opposite the largest side, the middle angle is opposite the middle side, and the smallest angle is
opposite the smallest side. Also, be very careful about order of operations!
Specifications: your program must
• ask the user to enter three side lengths in descending order (two consecutive equal sides is valid also), each of which can
be any positive number
order, or negative numbers, then your program does not need to work. You may also assume that the user enters
numbers that are the sidelengths of a real triangle! That is, you don't have to worry about the user entering 100, 2 and
1, because no real triangle has those three sidelengths (remember the triangle inequality?).
even one with decimals. You may assume that the user obeys if the user enters sides out of
• correctly calculate the three angles in the triangle, in degrees, under the assumptions ment ioned above.
• use the strategy I outlined above: Law of Cosines to find the largest angle, Law of Sines to find the middle angle, and
subtracting to find the smallest angle.
• print out the three angles in degrees, each on a separate line. You don't have to worry about how many decimal
places they are out put with.
that is, if they enter side lengths that are either not all positive, aren't entered in
Challenge: if the user is disobedient
descending order, or don't satisfy the triangle inequality - make the program print out a niessage saying so instead of trying
to perform the computations (which could result in an error). You might need to read ahead to do this.
3
MacBook Air
1)
F7
F9
F10
F11
F12

Expert Solution

Step 1 Code

#include <stdio.h>
#include <math.h>

#define RAD_2_DEG  (57.2958)
#define TOTAL_DEG  (180.0)

int main( void )
{

    int  a, b, c;   // user entered lengths of sides of triangle
    double A, B, C; // calculated angle values in degrees

    scanf( "%d %d %d", &a, &b, &c );


    // note: acos returns the angle in radians
    //       and one radian is (approx) 57.2958 degrees
    A = RAD_2_DEG * acos((double)(b*b + c*c - a*a)/(2.0*b*c));

    B = RAD_2_DEG * acos((double)(c*c + a*a - b*b)/(2.0*a*c));

    // third angle done this way to absorb fractional degree errors
    C = TOTAL_DEG -(A + B);

    printf("%.2f %.2f %.2f", A, B, C);
    return 0;
}

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions