
A channel 27m wide falls at a right angle into another channel 64m wide. The greatest length of the log that can be floated along this system of channels is
A. 120m
B. 125m
C. 100m
D. 110m
Answer
586.2k+ views
Hint: In this question, first we will make the diagram of the question. After this using a trigonometric function we will find the length of the log in terms of angle $\theta $. Now to find the maximum possible length of the log, we will differentiate the function which represents the length of log w.r.t $\theta $ and then equate it to zero. And finally use this result to find the length of the log.
Complete step-by-step answer:
The diagram of question is as follow:
It is give that”
AB channel is vertical channel while channel CD is horizontal and we have to find the length of log AD.
In $\vartriangle $ABO, we know that trigonometric function $\sin \theta $ is given by:
$\sin \theta $=$\dfrac{{{\text{perpendicular}}}}{{{\text{Hypotenuse}}}}$
$\therefore $ $\sin \theta $ =$\dfrac{{27}}{{OA}}$.
OA = $\dfrac{{27}}{{\sin \theta }}$
In $\vartriangle $COD, we know that trigonometric function $\cos \theta $:
$\therefore $ OD =$\dfrac{{64}}{{\cos \theta }}$.
Therefore, the length of log as function of angle $\theta $ is:
Length of log =AD = OA +OD=$\dfrac{{27}}{{\sin \theta }}$+$\dfrac{{64}}{{\cos \theta }}$=$f(\theta )$
To get the maximum length of log, we will differentiate the function $f(\theta )$w.r.t. $\theta $ and equate it to zero.
i.e. ${f'}(\theta )$=0
$ \Rightarrow \dfrac{{0 - 27\cos \theta }}{{{{\sin }^2}\theta }} + \dfrac{{0 - ( - 64\sin \theta )}}{{{{\cos }^2}\theta }}$ =0
\[ \Rightarrow \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }} = \dfrac{{27}}{{64}}\]
$ \Rightarrow {\tan ^3}\theta = \dfrac{{27}}{{64}}$
$ \Rightarrow \tan \theta = {\left( {\dfrac{{27}}{{64}}} \right)^{\dfrac{1}{3}}} = \dfrac{3}{4}$
From the above figure, we can say that:
$\sin \theta = \dfrac{3}{5}$ and $\cos \theta = \dfrac{4}{5}$ .
Therefore, the length of log = OA + OD = $\dfrac{{27}}{{\sin \theta }}$+$\dfrac{{64}}{{\cos \theta }}$=$\dfrac{{27}}{{\dfrac{3}{5}}}$m+$\dfrac{{64}}{{\dfrac{4}{5}}}$m= 125m
So, option B is the correct one.
Note- We should find the trigonometric ratio in a right triangle. The relation between the trigonometric function and sides of a right triangle is given as:
$\sin \theta = \dfrac{{{\text{perpendicular}}}}{{hypotenuse}},\cos \theta = \dfrac{{{\text{Base}}}}{{hypotenuse}},\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{Base}}}}$ . Before solving this question, you should also know how to find the maximum and minimum value of a function.
Complete step-by-step answer:
The diagram of question is as follow:
It is give that”
AB channel is vertical channel while channel CD is horizontal and we have to find the length of log AD.
In $\vartriangle $ABO, we know that trigonometric function $\sin \theta $ is given by:
$\sin \theta $=$\dfrac{{{\text{perpendicular}}}}{{{\text{Hypotenuse}}}}$
$\therefore $ $\sin \theta $ =$\dfrac{{27}}{{OA}}$.
OA = $\dfrac{{27}}{{\sin \theta }}$
In $\vartriangle $COD, we know that trigonometric function $\cos \theta $:
$\therefore $ OD =$\dfrac{{64}}{{\cos \theta }}$.
Therefore, the length of log as function of angle $\theta $ is:
Length of log =AD = OA +OD=$\dfrac{{27}}{{\sin \theta }}$+$\dfrac{{64}}{{\cos \theta }}$=$f(\theta )$
To get the maximum length of log, we will differentiate the function $f(\theta )$w.r.t. $\theta $ and equate it to zero.
i.e. ${f'}(\theta )$=0
$ \Rightarrow \dfrac{{0 - 27\cos \theta }}{{{{\sin }^2}\theta }} + \dfrac{{0 - ( - 64\sin \theta )}}{{{{\cos }^2}\theta }}$ =0
\[ \Rightarrow \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }} = \dfrac{{27}}{{64}}\]
$ \Rightarrow {\tan ^3}\theta = \dfrac{{27}}{{64}}$
$ \Rightarrow \tan \theta = {\left( {\dfrac{{27}}{{64}}} \right)^{\dfrac{1}{3}}} = \dfrac{3}{4}$
From the above figure, we can say that:
$\sin \theta = \dfrac{3}{5}$ and $\cos \theta = \dfrac{4}{5}$ .
Therefore, the length of log = OA + OD = $\dfrac{{27}}{{\sin \theta }}$+$\dfrac{{64}}{{\cos \theta }}$=$\dfrac{{27}}{{\dfrac{3}{5}}}$m+$\dfrac{{64}}{{\dfrac{4}{5}}}$m= 125m
So, option B is the correct one.
Note- We should find the trigonometric ratio in a right triangle. The relation between the trigonometric function and sides of a right triangle is given as:
$\sin \theta = \dfrac{{{\text{perpendicular}}}}{{hypotenuse}},\cos \theta = \dfrac{{{\text{Base}}}}{{hypotenuse}},\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{Base}}}}$ . Before solving this question, you should also know how to find the maximum and minimum value of a function.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

