Answer
Verified
457.8k+ views
Hint: In this particular question use the concept of cosine of angle in any triangle which is given as, $\cos \theta = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$ where a and b are the adjacent sides of that angle and c is the opposite side of that angle and later on differentiate this equation after simplifying w.r.t time so use these concepts to reach the solution of the question.
Complete step by step answer:
Two ships A and B are sailing straight away from a fixed point O along routes such that, $\angle AOB = {120^o}$.
Now it is given that OA = 8 km and OB = 6 km.
Let OA = a, OB = b, and AB = c Km as shown in the following figure.
Now it is also given that A is sailing at the rate of 20 km/hr. while the ship B sailed at the rate of 30 km/hr.
$ \Rightarrow \dfrac{{da}}{{dt}} = 20$ Km/hr. and $\dfrac{{db}}{{dt}} = 30$ Km/hr........................ (1)
Now we have to find out the distance between A and B is changing at the rate (in km/hr.) i.e. $\dfrac{{dc}}{{dt}}$
Now apply cosine rule in the above triangle we have,
$ \Rightarrow \cos \theta = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$, Where a and b are the adjacent sides of that angle and c is the opposite side of that angle.
$ \Rightarrow \cos {120^o} = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
Now as we know that $\cos {120^o} = \dfrac{{ - 1}}{2}$ so we have,
$ \Rightarrow - \dfrac{1}{2} = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
$ \Rightarrow {a^2} + {b^2} - {c^2} = - ab$
$ \Rightarrow {c^2} = {a^2} + {b^2} + ab$................. (2)
Now substitute the values we have,
$ \Rightarrow {c^2} = {8^2} + {6^2} + 8\left( 6 \right) = 64 + 36 + 48 = 148$
$ \Rightarrow c = \sqrt {148} $................ (3)
Now differentiate equation (2) w.r.t time (t) we have,
\[ \Rightarrow \dfrac{d}{{dt}}{c^2} = \dfrac{d}{{dt}}{a^2} + \dfrac{d}{{dt}}{b^2} + \dfrac{d}{{dt}}ab\]
Now as we know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ so we have,
\[ \Rightarrow 2c\dfrac{{dc}}{{dt}} = 2a\dfrac{{da}}{{dt}} + 2b\dfrac{{db}}{{dt}} + a\dfrac{{db}}{{dt}} + b\dfrac{{da}}{{dt}}\]
Now substitute all the values in the above equation we have,
\[ \Rightarrow 2\sqrt {148} \dfrac{{dc}}{{dt}} = 2\left( 8 \right)\left( {20} \right) + 2\left( 6 \right)\left( {30} \right) + 8\left( {30} \right) + 6\left( {20} \right)\]
Now simplify this we have,
\[ \Rightarrow 2\sqrt {148} \dfrac{{dc}}{{dt}} = 1040\]
\[ \Rightarrow \dfrac{{dc}}{{dt}} = \dfrac{{1040}}{{2\sqrt {148} }} = \dfrac{{520}}{{\sqrt {148} }} = \dfrac{{520}}{{2\sqrt {37} }} = \dfrac{{260}}{{\sqrt {37} }}\] Km/hr.
So this is the required answer.
Hence option (a) is the correct answer.
Note:
whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiating properties such as $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ and always recall the cosine rule in any triangle which is stated above, so use these properties as above applied we will get the required answer.
Complete step by step answer:
Two ships A and B are sailing straight away from a fixed point O along routes such that, $\angle AOB = {120^o}$.
Now it is given that OA = 8 km and OB = 6 km.
Let OA = a, OB = b, and AB = c Km as shown in the following figure.
Now it is also given that A is sailing at the rate of 20 km/hr. while the ship B sailed at the rate of 30 km/hr.
$ \Rightarrow \dfrac{{da}}{{dt}} = 20$ Km/hr. and $\dfrac{{db}}{{dt}} = 30$ Km/hr........................ (1)
Now we have to find out the distance between A and B is changing at the rate (in km/hr.) i.e. $\dfrac{{dc}}{{dt}}$
Now apply cosine rule in the above triangle we have,
$ \Rightarrow \cos \theta = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$, Where a and b are the adjacent sides of that angle and c is the opposite side of that angle.
$ \Rightarrow \cos {120^o} = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
Now as we know that $\cos {120^o} = \dfrac{{ - 1}}{2}$ so we have,
$ \Rightarrow - \dfrac{1}{2} = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
$ \Rightarrow {a^2} + {b^2} - {c^2} = - ab$
$ \Rightarrow {c^2} = {a^2} + {b^2} + ab$................. (2)
Now substitute the values we have,
$ \Rightarrow {c^2} = {8^2} + {6^2} + 8\left( 6 \right) = 64 + 36 + 48 = 148$
$ \Rightarrow c = \sqrt {148} $................ (3)
Now differentiate equation (2) w.r.t time (t) we have,
\[ \Rightarrow \dfrac{d}{{dt}}{c^2} = \dfrac{d}{{dt}}{a^2} + \dfrac{d}{{dt}}{b^2} + \dfrac{d}{{dt}}ab\]
Now as we know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ so we have,
\[ \Rightarrow 2c\dfrac{{dc}}{{dt}} = 2a\dfrac{{da}}{{dt}} + 2b\dfrac{{db}}{{dt}} + a\dfrac{{db}}{{dt}} + b\dfrac{{da}}{{dt}}\]
Now substitute all the values in the above equation we have,
\[ \Rightarrow 2\sqrt {148} \dfrac{{dc}}{{dt}} = 2\left( 8 \right)\left( {20} \right) + 2\left( 6 \right)\left( {30} \right) + 8\left( {30} \right) + 6\left( {20} \right)\]
Now simplify this we have,
\[ \Rightarrow 2\sqrt {148} \dfrac{{dc}}{{dt}} = 1040\]
\[ \Rightarrow \dfrac{{dc}}{{dt}} = \dfrac{{1040}}{{2\sqrt {148} }} = \dfrac{{520}}{{\sqrt {148} }} = \dfrac{{520}}{{2\sqrt {37} }} = \dfrac{{260}}{{\sqrt {37} }}\] Km/hr.
So this is the required answer.
Hence option (a) is the correct answer.
Note:
whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiating properties such as $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ and always recall the cosine rule in any triangle which is stated above, so use these properties as above applied we will get the required answer.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Which is the first open university in India A Andhra class 10 physics CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE