#include<stdio.h>
void main()
{
int row,column,add,sub,h,noofrows,noofcolumns,j,finish;
printf("how many rows do you want in your triangle\n");
scanf("%d",&noofrows);
noofcolumns=noofrows-1;
for(row=1;row<=noofrows-1;row=row+1)
{
for(column=-noofcolumns;column<=noofcolumns;column=column+1)
{
add=row+column;
sub=row-column;
if(add==1sub==1)
{
printf("*");
}
else
{
printf(" ");
}
}
printf("\n");
}
finish=((noofrows+noofrows)-1);
for(j=1;j<=finish;j=j+1)
{
printf("*");
}
scanf("%d",&h);
}
Nothing.The C language only recognizes a few keywords, like "for" and "if". Most of what's in a C program ... that doesn't reference routines in the C program itself ... are library calls, and cputs() is one of those. What it does is write its argument (which should be a pointer to a character string) to the console... console put string.
To draw a random triangle.... Declare six variables.... H1 and V1 H2 and V2 H3 and V3 Then using the random function apply the random to the height of your form - this will be H1 Do the same with the width of the form - this will be V1 -- do this twice more, for H2 and V2 - and then for H3 and V3.. Then draw a line from (H1, V1) to (H2, V2) Draw another line from (H1, V1) to (H3, V3) Draw the last line from (H2, V2) to (H3, V3) -- I can't guarantee it will look great, but it _will_ be a triangle. ==== To draw a specific triangle you need much more coding, involving sine rule an other trigonometry.
Write a C program to Draw a RAINBOW and fill the suitable colors ...
draw a line chart.
Nothing. (You can use this word as identifier though.)
Let's assume the triangle has points A, B, and C. Method 1 (3 lines) Draw two lines across the triangle parallel to line segment AB. Now you have two trapezoids and one triangle. Draw another line from C to the any point on the closest of the two lines you just drew, splitting the triangle into two more triangles. Method 2 (2 lines) Draw one line across the triangle parallel to line segment AB. Now you have one trapezoid and one triangle. Draw a second line that passes through C and is perpendicular to AB, splitting the trapezoid into two trapezoids and the triangle into 2 triangles. Method 3 (3 lines) Draw one line from point C to any point on line segment AB. Then draw a line parallel to AC and one parallel to BC, but don't let them cross the line you just drew.
An equilateral triangle MUST be acute. Draw a staright line AB. Take a compass and put its pin point at A and the pencil point at B. Draw an arc on one side of AB. Without disturbing the compass setting, move the pin point to B and draw another arc to intersect the first arc at C. Join CA and CB. Then ABC is an equilateral triangle.
a. (2,6). you draw it on a graph (x,y) b. c. d. make the triangle, a. lies between b. and c.
Yes, you can. You can pass an array of data to your console application.
Draw a circle using a compass. Then, without changing the compass setting, place its point on the circumference of the circle, at any point A, and draw two arcs to intersect the circumference at B and C. Move the compass to B and draw another arc to intersect the circumference at D; and then from C to E. ADE will be an inscribed equilateral triangle.
Draw a horizontal line AB equal to one of the side lengths. From A draw an arc of a circle of radius one of the remaining lengths. From B draw an arc of a circle of radius the third length. Where the arcs intersect is point C. Join AC and BC. Voila!
Side c of a triangle is opposite angle C
If the 13 is the longest side of that right triangle, then the missing side is 5 . If 'c' is the longest side of that right triangle, then the missing one is 17.692 (rounded).
a^2 + b^2 = c^2 c= hypotenuse a and b are the legs (sides) of the triangle
In right triangle ABC, angle C is a right angle, AB = 13and BC = 5 What is the length of AC? Draw the triangle to help visualize the problem.
Yes, it is quite simple.Draw a straight line segment, AB. Put the compass point at A and open it so that the pencil point is at B. Then draw an arc. Next, without changing the compass setting move it to B and draw another arc to cut the previous arc at C. [Actually there will be two points, one on either side of AB.] Using the straight edge, join AC and BC. Then ABC is an equilateral triangle.
If any one of them is longer than or equal to the sum of the other two, they can't form a triangle. If the lengths of the line segments are a, b and c, they form a triangle iff:a + b > ca + c > bb + c > a