Three boats met at point O, 1st boat went to north, 2nd went to east and 3rd went to south.The three boats sailed for 2 hours, when they stopped and arrived in points ABC, which make a right angle triangle.The distance between the 1st and 2nd boat is 60km.2nd boat sailed 6km/h faster than the first.
1)Find the distance between boat nr. 1 and boat nr 3.
2)After the two hour journey, boat nr.2 sailed for another hour, and arrived to the point D, find the angle of the point D.
A little background, this is not my homework, sort of.I am 14(no living in the UK nor in the native-english speaking countries, as you may have already noticed) and my math teacher gave me this extra math problem, she said that it was extremely hard and this kind of problems are solved in the college not in high school.So I gave it a shot and found it hard.My answer ist that the distance between nr. 1 boat and nr.3 boat is about 84,85 (km).And I have no idea how to calculate the second answer, or if the first answer is right..
I would really appreciate any help..Thank you in advance!
1)Find the distance between boat nr. 1 and boat nr 3.
2)After the two hour journey, boat nr.2 sailed for another hour, and arrived to the point D, find the angle of the point D.
A little background, this is not my homework, sort of.I am 14(no living in the UK nor in the native-english speaking countries, as you may have already noticed) and my math teacher gave me this extra math problem, she said that it was extremely hard and this kind of problems are solved in the college not in high school.So I gave it a shot and found it hard.My answer ist that the distance between nr. 1 boat and nr.3 boat is about 84,85 (km).And I have no idea how to calculate the second answer, or if the first answer is right..
I would really appreciate any help..Thank you in advance!
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assuming AO = x, BO = 12+x, AB = 60
so x^2 + (12+x)^2 = 60^2, x = 36
triangle ABC is right triangle, B is 90 degrees
triangles AOB is similar to BOC and ABC, BC = 80, AC = 100 (answer to question1)
BO = 48, speed of B = 24 km/h
OD = 72
denote angle ADO as @
tan @ = AO/OD = 36/72 = 1/2
@ = 26.6 degrees
denote angle ODC as B
tan B = 64/72
B = 41.6
angle D = @+B = 68.2 degrees
so x^2 + (12+x)^2 = 60^2, x = 36
triangle ABC is right triangle, B is 90 degrees
triangles AOB is similar to BOC and ABC, BC = 80, AC = 100 (answer to question1)
BO = 48, speed of B = 24 km/h
OD = 72
denote angle ADO as @
tan @ = AO/OD = 36/72 = 1/2
@ = 26.6 degrees
denote angle ODC as B
tan B = 64/72
B = 41.6
angle D = @+B = 68.2 degrees