I will assume that your previous post of
"FROM THE TOP OF A VERTICAL CLIFF 200M SEA LEVEL A BOAT WAS SIGHTED AT ANGLE OF 26CELSIUS ? "
and this post constitute one and the same question.
Secondly, I will assume that you meant an
"angle of 26 degrees" , not celsius, which is a measure of temperature.
Looks like tan26 = 200/x or
x = 200/tan26 = appr.410.1 m
b) so if the new distance is 410.1+1000 or 1410.1 m, let the new angle be theta
tan(theta) = 200/1410.1
theta = appr. 8.1 degrees
c) from b) we know the rate is 1000/(10/60) m/h or 6000 m/h
so in 45 minutes the boat would have gone another (45/60)(6000) = 4500 m
Add this to the origianl 410.1 and repeat steps of b)
A)HOW FAR IS THEBOAT FROM THE BASE OF THE CLIFF?
B) IF AFTER 10MIN THE BOAT HAS SAILED 1KM AWAY FROM THE CLIFF WHAT IS THE NEW ANGLE OF DEPRESSION OF THE BOAT ?
C) HOW FAR IS THE BOAT FROM THE BASE OF THE CLIFF AFTER 45MIN?
1 answer
1 answer