Reasoning :: Binary logic,Clock And Calenders :: Clock

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• Binary logic,Clock And Calenders - Clock
• Binary logic,Clock And Calenders - Calenders

Find the tim between 2 and 3 O'clock at which the minute hand and the hour hand are on the same straight line but are facing opposite directions.

Answer:

Explanation:

When two hands are pointing opposite direction and

are on a straight line the angle between them would be

$180^{o}$ and h=2.

$180^{o}=\frac{11}{2}m-30h$

$\frac{11}{2}m=180+60=240$

$m=\frac{480}{11}=43\frac{7}{11}$

So,at $43\frac{7}{11}$ min past $2$

the hands will be at $180^{o}$

Find the tim between 2 and 3 O'clock at which the minute hand and the hour hand are perpendicular to each other.

Answer:

Explanation:

When the two hands are perpendicular $\theta = 90^{o}$ and h=2

$\therefore \theta=$$(\frac{11}{2}m-30h)$ or $(30h-\frac{11}{2}m)$

$90=\frac{11}{2}m-30\times 2$

$\frac{11}{2}m=150$

$m=\frac{300}{11}$

$27\frac{3}{11}$ min past $2$.

or $90=30 \times 2-\frac{11}{2}$

$\frac{11}{2}=-30$

As m cannot be negative,this case is not possible.

so, the hands are perpendicular to each other only once ie, at $27\frac{3}{11}$ min past $2$.

Find the tim between 2 and 3 O'clock at which the minute hand and the hour hand overlap

Answer:

Explanation:

When the two hands overlap, the angle between them is $0^{o}$.

$\theta = \mid \frac{11}{2}m-30h \mid$

$\therefore \theta=0^{o}$ and $h=2$

$\frac{11}{2}m=30 \times 2$

$m= \frac{120}{11}=10\frac{10}{11}$min past ${2}$

Find the tim between 2 and 3 O'clock at which the minute hand and the hour hand make an angle of $60^{o}$ with each other.

Answer:

Explanation:

In formula $\theta = \mid \frac{11}{2} m-30h \mid$

$\theta=60^{o}$ and h=2

$\therefore 60=\frac{11}{2}m-30 \times 2$

$\frac{11}{2}m=120$

$m=\frac{240}{11}=21\frac{9}{11}$min past $2$

or $60=30\times 2-\frac{11}{2}m$

$\therefore \frac{11}{2} m=0$

m=0

Therefore, the angle between the hour hand and the minute hand is $60^{o}$

at 2 o'clock and at $21\frac{9}{11}$minutes past $2 o'clock$

What is the angle between the minute and the hour hand of a clock at 3 hours 40 minutes?

Answer:

Explanation:

The angle between the hands can be calculated by $\theta=$$\mid \frac{11}{2} m 4-30h  \mid$ , where m is minutes and h is hour.

Here m= 40 and h=3

$\therefore \theta=\mid \frac{11}{2}\times - 30 \times 3 \mid$

$=\mid 200-90 \mid$

=$130^{o}$

• Binary logic,Clock And Calenders - Binary logic
• Binary logic,Clock And Calenders - Clock
• Binary logic,Clock And Calenders - Calenders