As observed from the top of a light house,100m above the sea level,the angle of depression of a ship sailing directly towards it,changes from 30 degree to 45 degree.Determine the distance travelled by the ship during this period
nazilmohammad:
Pls answer as soon as possible I have exam tomorrow board exam 10th stndarad
standard
Tomorrow important exam not board exam
Answers
Answered by
50
height of light housr=100m
let distance between two ships = x m
height of sheep=100m
near boat makes 45 degree and away boat makes 30 degree with top of light house
also let distance of nearer ship from foot of light house=y m
DRAW DIAGRAM IN YOUR NOTEBOOK FIRST
tan45=100/y
1=100/y
y=100m
tan 30= 100/(x+y)
1/√3=100/(x+100)
x+100=100√3
x=100√3-100
x=100(√3-1)m
So DISTANCE BETWEN SHIPS IS 100(√3-1)m
MARK AS BRAINLIEST ANSWER IF GOT IT
let distance between two ships = x m
height of sheep=100m
near boat makes 45 degree and away boat makes 30 degree with top of light house
also let distance of nearer ship from foot of light house=y m
DRAW DIAGRAM IN YOUR NOTEBOOK FIRST
tan45=100/y
1=100/y
y=100m
tan 30= 100/(x+y)
1/√3=100/(x+100)
x+100=100√3
x=100√3-100
x=100(√3-1)m
So DISTANCE BETWEN SHIPS IS 100(√3-1)m
MARK AS BRAINLIEST ANSWER IF GOT IT
MARK AS BRAINLIEST IF YOU GOT THE ANSWER
I can't mark brainlisest if two people did not write it won't let me
x=-100+100 root 3 so the Ans would be root 3 right???
OK DEAR , THANKS FOR SHARING INFORMATION ABOUT IT
NO, 100 SHOULD BE TAKE COMMON IN SECOND LAST STEP
Answered by
9
Answer :
There are two ways to solve this question -
1. Method 1
IN ΔABC ,
Tan 45 degree = AB/BC
1 = 100/BC
BC = 100
IN ΔABD ,
Tan 30 degree = AB/DB
1/√3 = AB/BD
1/√3 = 100/Y (Y = DB)
1/√3 = 100/100+X
100+X = 100√3
X = 100√3 - 100
X = 100 (√3 - 1)
METHOD 2:
IN ΔABC ,
Tan 45 degree = AB/BC
1 = 100/BC
BC = 100
IN ΔABD ,
Tan 30 degree = AB/DB
1/√3 = AB/BD
1/√3 = 100/Y (Y = DB)
Y = 100√3
Y = 100 * 1.732
Y = 173.2
DC = DB - BC
DC = 173.2 - 100
DC = 73.2 METERS.
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