triangle determine

[Thephantom23]

can u help me this problem :

Input :
4 3 5
output:
right triangle

input :
4 4 5
output:
isoscles triangle

input:
4 4 4
output:
equilateral triangle

can u help me for this ????/

[shadder]

Hi,

An effort from your side would be lovely.... could you show us what you did/tried?what problems are you facing?

[ephraimr]

Let three variables a,b,c be the lengths. Take this value from user. Now use some if conditions to determine the result. Where exactly is the problem?

[ephraimr]

Submit what you already wrote inside the code tags.

[boost lexical cast]

Let's say the 3 sides are a, b, c.

Right triangle: Any of the 2 sides ([a,b], [b,c], [a,c]) squared and added equals the square of the 3rd.
Example: a^2 + b^2 = c^2 so sqrt(a^2 + b^2) = c.

Isosceles triangle: 2 of the sides are equal and one is different from the other 2.

Equilateral triangle: All of the sides are equal.

You should be able to check for these conditions easily enough.

[b]EDIT[/b]
Here's code if you don't want to do it:

#include <iostream>
#include <cmath>
#include <conio.h> // Optional. For getch(); which pauses program
int main()
{
    int a, b, c;
    std::cout << "Enter 3 sides of triangle\n";
    std::cout << "1st side: ";
    std::cin >> a;
    std::cout << "2nd side: ";
    std::cin >> b;
    std::cout << "3rd side: ";
    std::cin >> c;
    std::cout << '\n';
    if ((a == b) && (b == c) && (a == c))
    {
        std::cout << "This triangle is an equilateral triangle\n";
    }
    if (((a == b) && ((a != c) || (b != c))) || ((a == c) && ((a != b) || (c != b))))
    {
        std::cout << "This triangle is an isosceles triangle\n";
    }
    if (((std::sqrt (a * a + b * b)) == c) || ((sqrt (b * b + c * c)) == a) || ((sqrt (a * a + c * c)) == b))
    {
        std::cout << "This triangle is a right triangle\n";
    }
    getch(); // Optional. Pauses program.
    return 0;
}

[boost lexical cast]

@gentleguy

Mistake was thinking on the wrong train of thought. My code doesn't use arrays.

[AterVulpes]

            A = acos((b*b+c*c-a*a)/(2*b*c));
            B = acos((a*a+c*c-b*b)/(2*a*c));
            C = acos((a*a+b*b-c*c)/(2*a*b));

Why on Earth does this result in a 270[°] triangle O_o ? (for 3,4,5)

Whole code:

#include <iostream>
#include <math.h>

const double M_PI = 4*atan(1);

int main()
{
    char loop;

    do
        {
        int a, b, c; //assuming whole numbers
        double A, B, C; //angles
        std::cout << "Input triangle side lengths\n";
        std::cin >> a;
        std::cin >> b;
        std::cin >> c;

        if ((a==b) & (b==c))
        {
            std::cout << "Equilateral triangle.\nAngles are: A=B=A=60[°]";
        }
        else if (((a==b) & (a!=c)) | ((a==c) & (b!=c)))
        {
            A = acos((b*b+c*c-a*a)/(2*b*c));
            B = acos((a*a+c*c-b*b)/(2*a*c));
            C = acos((a*a+b*b-c*c)/(2*a*b));

            std::cout << "Isoscles triangle. Angles are: A=" << A << "[rad] B=" << B << "[rad] C="  << C << "[rad]\n";;
        }
        else if ((a*a==b*b+c*c) | (b*b==a*a+c*c) | (c*c==a*a+b*b))
        {

            A = acos((b*b+c*c-a*a)/(2*b*c));
            B = acos((a*a+c*c-b*b)/(2*a*c));
            C = acos((a*a+b*b-c*c)/(2*a*b));

            std::cout << "Right triangle. Angles are: A=" << A << "[rad] B=" << B << "[rad] C="  << C << "[rad]\n";
        }
        else
        {
            A = acos((b*b+c*c-a*a)/(2*b*c));
            B = acos((a*a+c*c-b*b)/(2*a*c));
            C = acos((a*a+b*b-c*c)/(2*a*b));

            std::cout << "Scalene triangle.Angles are: A=" << A << "[rad] B=" << B << "[rad] C="  << C << "[rad]\n";
        }

        std::cout << "\n\nEvaluate another triangle y/n?\n";
        std::cin >> loop;
        loop = toupper(loop);
        }
        while (loop == 'Y');

    return 0;

}

[keskiverto]

If inputs are int, then there is no need for floating-point math (e.g. acos and sqrt) and right angled isosceles triangle will not occur.

[48T7M4Gy]

#include <iostream>
#include <cmath>

int main()
{
    double a = 0, b = 0, c = 0;     // sides
    double aa = 0, ab = 0, ac = 0;  // angles
    double p = 0;   // perimeter
    
    std::cout << "Please enter 3 sides: ";
    std::cin >> a >> b >> c;
    
    ac = acos( (a*a + b*b - c*c)/(2*a*b) );
    aa = acos( (b*b + c*c - a*a)/(2*a*b) );
    ab = M_1_PI - ac - aa;
    
    p = a + b + c;
    
    if( aa == M_PI/2 or ab == M_PI/2 or ac == M_PI/2)
        std::cout << "Right angled triangle\n";
    else if( p / 3 == a )
        std::cout << "Equilateral triange\n";
    else if ( p - a == 2*b or p - b == 2*a or p - c == 2*a)
        std::cout << "Isosceles triangle\n";
    else
        std::cout << "Scalene triangle\n";
}

[JLBorges]

void swap( int& a, int& b ) { const int temp = a ; a = b ; b = temp ; }

void make_ascending( int& a, int& b ) { if( b < a ) swap(a,b) ; }

void make_ascending( int& a, int& b, int& c ) { make_ascending(a,b) ; make_ascending(b,c) ; make_ascending(a,b) ; }

bool is_side( int a ) { return a > 0 ; }

bool is_triangle( int a, int b, int c ) // a, b, c in ascending order
{ return is_side(a) && is_side(b) && is_side(c) && (a+b) > c ;  }

enum type { isosceles, equilateral, right_angled, scalene_not_right_angled, not_a_triangle = -1 };

type triangle_type( int a, int b, int c )
{
    make_ascending(a,b,c) ;

    if( is_triangle(a,b,c) )
    {
        if( a==b ) return b==c ? equilateral : isosceles ;
        else return ( a*a + b*b == c*c ) ? right_angled : scalene_not_right_angled ;
    }
    else return not_a_triangle ;
}

[48T7M4Gy]

#include <iostream>

int main()
{
    double a = 0, b = 0, c = 0;
    
    std::cout << "Please enter 3 sides: ";
    std::cin >> a >> b >> c;
    
    double a2 = a * a, b2 = b * b, c2 = c * c, p = a + b + c;
    
    if( a == b and b == c)
        std::cout << "Equilateral triange\n";
    else if ( p - a == 2 * b or p - b == 2 * a or p - c == 2 * a)
        std::cout << "Isosceles triangle\n";
    else if ( a2 == b2 + c2 or b2 == a2 + c2 or c2 == a2 + b2)
        std::cout << "Right angled triangle\n";
    else
        std::cout << "Scalene triangle\n";
}

[AterVulpes]

http://cplusplus.happycodings.com/mathematics/code1.html