# Measure of angle formed by minute and hour hands

• Nov 6th 2010, 03:31 PM
Hellbent
Measure of angle formed by minute and hour hands
Hi,

What is the measure of the angle formed by the minute and hour hands of a clock at 1:50?

(A) $\displaystyle 90^o$
(B) $\displaystyle 95^o$
(C) $\displaystyle 105^o$
(D) $\displaystyle 115^o$
(E) $\displaystyle 120^o$

My guess was $\displaystyle 90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type.
I seek a better approach to this question and an explanation of this approach. I ask kindly.
• Nov 6th 2010, 03:43 PM
Quote:

Originally Posted by Hellbent
Hi,

What is the measure of the angle formed by the minute and hour hands of a clock at 1:50?

(A) $\displaystyle 90^o$
(B) $\displaystyle 95^o$
(C) $\displaystyle 105^o$
(D) $\displaystyle 115^o$
(E) $\displaystyle 120^o$

My guess was $\displaystyle 90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type.
I seek a better approach to this question and an explanation of this approach. I ask kindly.

A 12-hour clock has 12 subdivisions, each of which is $\displaystyle 30^o$

At 1:50, the minute hand is pointing at 10, so it is $\displaystyle (2)30^0$ left of 12.

The hour hand will have moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12.

This leaves it $\displaystyle 30^0+\left(\frac{5}{6}\right)30^0$ to the right of 12.
• Nov 6th 2010, 04:29 PM
Hellbent
Quote:

Originally Posted by Archie Meade
A 12-hour clock has 12 subdivisions, each of which is $\displaystyle 30^o$

At 1:50, the minute hand is pointing at 10, so it is $\displaystyle (2)30^0$ left of 12.

The hour hand will have moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12.

This leaves it $\displaystyle 30^0+\left(\frac{5}{6}\right)30^0$ to the right of 12.

My understanding is much better. I am having a problem with this part: The hour hand will have moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^o$, when the minute hand was pointing at 12.

Wouldn't it have been the minute hand that moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^o$? Seeing that it has moved from 12 to 10 - $\displaystyle 300^o$.
• Nov 6th 2010, 04:36 PM
Quote:

Originally Posted by Hellbent
My understanding is much better. I am having a problem with this part: The hour hand will have moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^o$, when the minute hand was pointing at 12.

Wouldn't it have been the minute hand that moved $\displaystyle \frac{10}{12}$ of $\displaystyle 30^o$? Seeing that it has moved from 12 to 10 - $\displaystyle 300^o$.

I didn't write the first post too well.

Imagine the time is initially 1 o'clock. The minute hand is at 12 and the hour hand is at 1.
Both hands move and the clock reads 1:50.

Yes, the minute hand will have moved through $\displaystyle 30^o$ ten times

while the hour hand will have moved through $\displaystyle \frac{30^o}{12}$ ten times.

The hour hand moves through $\displaystyle 30^0$ for a $\displaystyle 360^0$ movement of the minute hand.

I'm getting (D), not (C).
• Nov 6th 2010, 05:25 PM
Hellbent
Thanks.

Sorry, typo, it is indeed (D.) $\displaystyle 115^o$.

I just changed the values in this question to check my understanding:
What is the measure of the angle formed by the minute and hour hands of a clock at 3:45?
$\displaystyle \frac{45}{60} = \frac{3}{4}$

$\displaystyle 90^o + (\frac{3}{4})30^o = 112.5^o$ Then adding another $\displaystyle 90^o$ (intervening angle between 9 and 12) gives $\displaystyle 202.5^o$

Would the approach be the same for non-multiples of 5. Say, 4:47?
• Nov 6th 2010, 05:37 PM
Quote:

Originally Posted by Hellbent
Thanks.

Sorry, typo, it is indeed (D.) $\displaystyle 115^o$.

I just changed the values in this question to check my understanding:
What is the measure of the angle formed by the minute and hour hands of a clock at 3:45?
$\displaystyle \frac{45}{60} = \frac{3}{4}$

$\displaystyle 90^o + (\frac{3}{4})30^o = 112.5^o$ Then adding another $\displaystyle 90^o$ (intervening angle between 9 and 12) gives $\displaystyle 202.5^o$

Would the approach be the same for non-multiples of 5. Say, 4:47?

Yes,

Starting at 4 o'clock, the minute hand will swing through $\displaystyle \frac{47}{60}360^o$

Then the hour hand will swing through $\displaystyle \frac{47}{60}30^o$

You can simply think in terms of "fractions of an hour" and so "fractions of $\displaystyle 360^o$" and "fractions of $\displaystyle 30^o$"
• Nov 6th 2010, 05:41 PM
Hellbent
Thanks.