Question

In the summer term sal wlks to school starting soon after 7 am when the hour hand is at angle of 110 degrees to the minute hand. She arrives at school before 8 am when

Answer 1

Answer:

She walked for 46 minutes

Step-by-step explanation:

Given

When she leaves home

$h = 7am$

$\theta_1 = 110^o$

When she arrives at school

$H = 8am$

$\theta_2 = 110^o$

Required

Time to walk to school

First, we convert $\theta_1 = 110^o$ to minutes:

$\theta_1 = \frac{110}{6}$

$\theta_1 = 18.33$

$\theta_1 = 18$ --- approximated

From the question, we understand that she left home past 7.

This means that the difference between the hour hand (at 7) and the minutes hand is 18 minutes.

Also, the minute hand is between 0 and 30 minutes

Subtract 18 minutes from 30 minutes;

$Mins = 30-18=12$

She left home at 7:12am

Similarly;

$\theta_2 = \theta_1 = 18\ mins$

From the question, we understand that she arrived before 8.

This means that the difference between the hour hand (at 8) and the minutes hand is 18 minutes

Also, the minute hand is between 30 and 60 minutes mark.

At the 8hour hand, the minutes is 40

Add 18 minutes from 40 minutes;

$Mins = 40 + 18 = 58$

She arrived at 7:58am

The difference (d) is:

$d =7:58am - 7:12am$

$d =46$