The hour hand on my antique schoolhouse clock is 3 inches long and the minute hand is 4.5 inches long. What is the distance at 4 o'clock?

Extend the line of the minute hand down towards the number '6'.

Now add a line at right angles to this extension to meet the end of the hour hand.

You now have a right-angle triangle consisting of the two new line and the hour hand. The hour hand at 4 o'clock is at an angle of

60 degrees from the vertical.

The tip of the hour hand is 3*Cos(60) inches below the centre of the clock dial and 3*Sin(60) to the right if the clocks centre line.

Now construct triangle consisting of the vertical line from the minute hand tip to the line that was added previously to the end of the hour hand. That vertical line and the horizontal line added earlier form two sides of a larger right angle. The hypotenuse is the answer to this problem.

Using Pythagoras we get

the distance between the ends of the hands 'd' is given by:

d^2 = (4.5 + 3Cos(60))^2 + (3Sin(60))^2

= 36 + 6.75 = 42.75

So d = root(42.75) = 6.538 inches (to 3 decimal places).

I hope this helps.

distance d

d^2=a^2+b^2-2abcosx( where a hour hand ,b=min hand,x angle beweena,b),rule

d^2=3^2+(4.5)^2-2(3)(4.5)cos120

= 9+20.25-6(4.5)(-cos60)

=29.25 -27(- 1/2)

29.25+13.5=42.75

d=sqr(42.75=approximately6.54

d² = 3.5² + 4² - 2 x 3.5 x 4 cos 120⁰

d² = 3.5² + 4² - 2 x 3.5 x 4 cos 120⁰

d = 6.5 ins