Find triangle is equilateral,isosceles or right angled Code in C Language

Write a C program to Find triangle is equilateral,isosceles or right angled by accepting  the length of three sides of a triangle from input and then test and print the type of triangle.

Program will test the type of triangle is equilateral, isosceles, right angled or none of these and print the output to screen.Program will use the basic mathematics formulas of trigonometric to test type of triangle.

Find triangle is equilateral,isosceles or right angled code

#include “stdio.h”
void main()
{
int i, j; 
int temp; 
int sides[3]; 

int type; 
int isRightAngled; 
clrscr();

printf(“Please enter the three sides of triange\n”);

for(i = 0; i < 3; i++)
{
scanf(“%d”, &sides[i]);
}

for(i = 0; i < 3; i++)
{
for(j = i + 1; j < 3; j++)
{

if(sides[i] > sides[j])
{

temp = sides[i];
sides[i] = sides[j];
sides[j] = temp;
}
}
}

type = 3;
isRightAngled = 0;

if(sides[2] > sides[0] + sides[1]) /*check if triangle is valid */
type = 0;
else if(sides[0] == sides[2]) 
type = 1; 
else
{
if(sides[0] == sides[1]) 
type = 2; 

if(sides[2] * sides[2] == sides[0] * sides[0] + sides[1] * sides[1])
isRightAngled = 1;
}

switch(type)
{
case 0:
printf(“The triangle is Invalid\n”);
break;
case 1:
printf(“The triangle is Equilateral Triangle\n”);
break;
case 2:
if(isRightAngled == 1)
printf(“The triangle is Isosceles and Right Angled Triangle\n”);
else
printf(“The triangle is Isosceles Triangle\n”);
break;
case 3:
if(isRightAngled == 1)
printf(“The triangle is Scalene and Right Angled Triangle\n”);
else
printf(“The triangle is Scalene Triangle\n”);
break;

}
getch();
}

Display numbers as English word in C Language

Write a Display numbers as English word program in C to read an integer and display each of the digit of the integer in English.User has to provide a number which will be printed as English word on the screen.

You can take an example of 2345 will be printed as two three four five.

Display numbers as English word in C Code

#include <stdio.h>

void main ()
{
int number, digit1, digit2, digit3, digit4, digit5;
clrscr();
printf(“Enter a number (max 5 digits): “);
scanf(“%i”, &number);

if (number < 100000)
{
digit1 = number % 100000 / 10000;

switch (digit1)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit2 = number % 10000 / 1000;

switch (digit2)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit3 = number % 1000 / 100;

switch (digit3)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit4 = number % 100 / 10;

switch (digit4)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit5 = number % 10;

switch (digit5)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}
}
else
{
printf(“Entered Number is too long.”);
}
getch();
}

Fibonacci numbers program in C Language

Write a Fibonacci numbers program in C program in C to generate first 20 Fibonacci numbers using simple Code. User needs to enter till which number Fibonacci numbers should be printed through user Interface.

After getting the right input  program will print the  Fibonacci numbers .

The first 21 Fibonacci numbers F_n for n = 0, 1, 2, …, 20.By definition of Fibonacci numbers, it starts with first two numbers in the Fibonacci sequence as 0 and 1, and each subsequent number is the sum of the previous two.

like 0,1,1,2………

Fibonacci numbers program in C Code

#include<stdio.h>
#include<conio.h>

void main()
{
int a,b,c,num,i;
a=0,b=1,c=0;
clrscr();
printf(“\nEnter the number till which u want the series::”);
scanf(“%d”,&num);
printf(“\n”);

for(i=0;i<num;i++,c=a+b)
{
a=b;
b=c;
printf(“%d\t”,c);
}
getch();
}

Roots of a quadratic equation in C Language

Write a Roots of a quadratic equation program in C to compute the roots of a quadratic equation.User has to provide values of variables of equation initially.

Then program will find the value of Roots of a quadratic equation .

Roots of a quadratic equation Code

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

void main()
{
float a,b,c,root1,root2;
clrscr();
printf(“\n Enter values of a,b,c for finding roots of a quadratic eq:\n”);
scanf(“%f%f%f”,&a,&b,&c);

if(b*b>4*a*c)
{
root1=-b+sqrt(b*b-4*a*c)/2*a;
root2=-b-sqrt(b*b-4*a*c)/2*a;
printf(“\n*****ROOTS ARE*****\n”);
printf(“\n root1=%f\n root2=%f”,root1,root2);
}
else
printf(“\n Imaginary Roots.”);
getch();
}