Mixing
A man on the top of the tower, standing on the sea.
A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
My Attempt,
We know, Speed $=frac {dist. }{time }$
$=frac {CD}{10}$.
Also, $Tan 60=frac {AB}{BC}$
$sqrt {3}=frac {AB}{BC}$
$AB=sqrt {3} BC$.
Now, what do I have to do further?
trigonometry
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
add a comment |
A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
My Attempt,
We know, Speed $=frac {dist. }{time }$
$=frac {CD}{10}$.
Also, $Tan 60=frac {AB}{BC}$
$sqrt {3}=frac {AB}{BC}$
$AB=sqrt {3} BC$.
Now, what do I have to do further?
trigonometry
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18
add a comment |
A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
My Attempt,
We know, Speed $=frac {dist. }{time }$
$=frac {CD}{10}$.
Also, $Tan 60=frac {AB}{BC}$
$sqrt {3}=frac {AB}{BC}$
$AB=sqrt {3} BC$.
Now, what do I have to do further?
trigonometry
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
My Attempt,
We know, Speed $=frac {dist. }{time }$
$=frac {CD}{10}$.
Also, $Tan 60=frac {AB}{BC}$
$sqrt {3}=frac {AB}{BC}$
$AB=sqrt {3} BC$.
Now, what do I have to do further?
trigonometry
trigonometry
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
asked Jan 25 '17 at 15:02
blue_eyed_...
3,23821645
3,23821645
You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18
add a comment |
You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18
You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18
You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18
add a comment |
2 Answers
2
active
oldest
votes
If the side length of $CB $ is $x $, then in $triangle ABC $, we have length of $AB =sqrt {3}x $. Now in $triangle ABD $, we have $BD =sqrt {3}x cot 30 =3x $. Thus $CD=2x $.
If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps. 
edited Nov 21 '18 at 6:13
Community♦
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
add a comment |
Let's consider $AB$ to be $h$. Then,
$$
BD=frac{h}{tan30^circ}=hsqrt{3} \
BC=frac{h}{tan60^circ}=frac{h}{sqrt{3}}\
CD=BD-BC=frac{2h}{sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=frac{2h}{sqrt{3}}=600x
implies frac{h}{xsqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
frac{BC}{x}=frac{h}{xsqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If the side length of $CB $ is $x $, then in $triangle ABC $, we have length of $AB =sqrt {3}x $. Now in $triangle ABD $, we have $BD =sqrt {3}x cot 30 =3x $. Thus $CD=2x $.
If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps.
edited Nov 21 '18 at 6:13
Community♦
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
add a comment |
If the side length of $CB $ is $x $, then in $triangle ABC $, we have length of $AB =sqrt {3}x $. Now in $triangle ABD $, we have $BD =sqrt {3}x cot 30 =3x $. Thus $CD=2x $.
If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps.
edited Nov 21 '18 at 6:13
Community♦
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
add a comment |
If the side length of $CB $ is $x $, then in $triangle ABC $, we have length of $AB =sqrt {3}x $. Now in $triangle ABD $, we have $BD =sqrt {3}x cot 30 =3x $. Thus $CD=2x $.
If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps.
edited Nov 21 '18 at 6:13
Community♦
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
If the side length of $CB $ is $x $, then in $triangle ABC $, we have length of $AB =sqrt {3}x $. Now in $triangle ABD $, we have $BD =sqrt {3}x cot 30 =3x $. Thus $CD=2x $.
If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps.
edited Nov 21 '18 at 6:13
Community♦
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
edited Nov 21 '18 at 6:13
Community♦
1
edited Nov 21 '18 at 6:13
Community♦
1
edited Nov 21 '18 at 6:13
Community♦
1
1
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
answered Jan 25 '17 at 15:07
Rohan
27.7k42444
27.7k42444
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
add a comment |
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
what is $triangle BD$??
– blue_eyed_...
Jan 25 '17 at 15:12
add a comment |
Let's consider $AB$ to be $h$. Then,
$$
BD=frac{h}{tan30^circ}=hsqrt{3} \
BC=frac{h}{tan60^circ}=frac{h}{sqrt{3}}\
CD=BD-BC=frac{2h}{sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=frac{2h}{sqrt{3}}=600x
implies frac{h}{xsqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
frac{BC}{x}=frac{h}{xsqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
add a comment |
Let's consider $AB$ to be $h$. Then,
$$
BD=frac{h}{tan30^circ}=hsqrt{3} \
BC=frac{h}{tan60^circ}=frac{h}{sqrt{3}}\
CD=BD-BC=frac{2h}{sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=frac{2h}{sqrt{3}}=600x
implies frac{h}{xsqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
frac{BC}{x}=frac{h}{xsqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
add a comment |
Let's consider $AB$ to be $h$. Then,
$$
BD=frac{h}{tan30^circ}=hsqrt{3} \
BC=frac{h}{tan60^circ}=frac{h}{sqrt{3}}\
CD=BD-BC=frac{2h}{sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=frac{2h}{sqrt{3}}=600x
implies frac{h}{xsqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
frac{BC}{x}=frac{h}{xsqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
Let's consider $AB$ to be $h$. Then,
$$
BD=frac{h}{tan30^circ}=hsqrt{3} \
BC=frac{h}{tan60^circ}=frac{h}{sqrt{3}}\
CD=BD-BC=frac{2h}{sqrt{3}}
$$
If the speed of the boat is $x$ metres per second,
$$
CD=frac{2h}{sqrt{3}}=600x
implies frac{h}{xsqrt{3}}=300
$$
Time in which the boat travels distance $BC$ is :
$$
frac{BC}{x}=frac{h}{xsqrt{3}}=300
$$
Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
answered Jan 25 '17 at 15:15
Vishnu V.S
715210
715210
add a comment |
add a comment |
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You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate
– Henry
Jan 25 '17 at 15:18