two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. the angles of depression of two ships as observed from the top of the light house are 60 degree and 45 degree. if the height of the light house is 200 m , find the distance between the two ships.
Answers
For diagram refer to the attachment!!
★ Given that
- According to the given question, ships (B and D) on either side of the light house (AC = 200 m).
- The angle of the depression of two ships observed are ∠ABC = 60° and ∠ADC = 45°.
★ To Find
- the distance between the two ships (i.e. BD).
★ Solution
As we know that
In the ∆ABC right angled at C,
→ Tan 60° = AC/BC
→ √3 = 200/BC
Therefore, BC = 200/√3 m
In the ∆ACD right angled at C,
→ Tan 45° = AC/CD
→ 1 = 200/CD
Therefore, CD = 200 m
Now,
☞ BD = BC + CD
☞ BD = 200/√3 + 200
☞ BD = 200(1+√3)/√3
We also know that, √3 = 1.73
Then,
☞ BD = 200(1+1.73)/ 1.73
☞ BD = 54600/173
☞ BD = 315.60 m (approx)
which is your required answer!!
So, the distance between the two ships is 315.60 m.
Answer:
★ Given that
According to the given question, ships (B and D) on either side of the light house (AC = 200 m).
The angle of the depression of two ships observed are ∠ABC = 60° and ∠ADC = 45°.
★ To Find
the distance between the two ships (i.e. BD).
★ Solution
As we know that
In the ∆ABC right angled at C,
→ Tan 60° = AC/BC
→ √3 = 200/BC
Therefore, BC = 200/√3 m
In the ∆ACD right angled at C,
→ Tan 45° = AC/CD
→ 1 = 200/CD
Therefore, CD = 200 m
Now,
☞ BD = BC + CD
☞ BD = 200/√3 + 200
☞ BD = 200(1+√3)/√3
We also know that, √3 = 1.73
Then,
☞ BD = 200(1+1.73)/ 1.73
☞ BD = 54600/173
☞ BD = 315.60 m (approx)
which is your required answer!!
So, the distance between the two ships is 315.60 m.