A knight wanted to visit a princess. He had to arrive at exactly 17h00.
If he travelled at 15 km per hour, he would arrive one hour too early.
If he travelled at 10 km per hour, he would arrive one hour too late.
Questions : 1- At what time did he leave?
2- What distance did he travel?
3- At what speed did he travel?

He had to arrive at 5:00pm

If he traveled at 15km per hour, he would arrive one hour too early

If he traveled at 10km per hour, he would arrive one hour too late.

At what time did he leave, what distance did he travel, and at what speed did he travel

:

Let d = distance traveled

:

Write an time equation; time = dist/speed

Travel time at 10 km/hr - travel time at 15 km/h = 2 hrs

d%2F10 - d%2F15 = 2

Multiply by 30

3d - 2d = 30(2)

d = 60 km is the distance

:

Find the travel time at 15 km/hr

60%2F15 = 4 hrs

:

let s = speed required to get there on time

:

required travel time - 15km/hr travel time = 1

60%2Fs - 4 = 1 hr

60 - 4s = 1s

60 = 1s + 4s

60 = 5s

s = 12 km/hr to arrive on time

:

Required travel time: 60%2F12 = 5 hrs

He would have to leave at noon

;

Summarize, depart at noon, travel for 5 hrs at 12 km/hr, a dist of 60 km

Step-by-step explanation:

answer

step-by-step explanation:

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92