Question

To Prove: The elevation of hour hand at any time in degrees in 'analogue clock' is —
$((h \times 30) + (m \times 0.5)) ^{o}$
where, h stands for hours and m stands for minutes of current time.

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Answer 1

Explanation:

C program to find angle between hour and minute hands

#include <stdio.h>

#include <stdlib.h>

// Utility function to find minimum of two integers

int min(int x, int y) { return (x < y)? x: y; }

int calcAngle(double h, double m)

{

// validate the input

if (h <0 || m < 0 || h >12 || m > 60)

printf("Wrong input");

if (h == 12) h = 0;

if (m == 60)

{

m = 0;

h += 1;

if(h>12)

h = h-12;

}

// Calculate the angles moved by hour and minute hands

// with reference to 12:00

int hour_angle = 0.5 * (h*60 + m);

int minute_angle = 6*m;

// Find the difference between two angles

int angle = abs(hour_angle - minute_angle);

// Return the smaller angle of two possible angles

angle = min(360-angle, angle);

return angle;

}

// Driver Code

int main()

{

printf("%d n", calcAngle(9, 60));

printf("%d n", calcAngle(3, 30));

return 0

}