Question

A boat has to cross a river. It crosses the river by making an angle of \[60^\circ \] with the bank of the river due to the stream of the river, and travels a distance of 600 m to reach the other side of the river. What is the width of the river?

Answer 1

Hint: Here, we need to find the width of the river. We will use the formula for the trigonometric ratio, sine of an angle, in a right angled triangle to form an equation. We will solve the equation to find the width of the river.

Formula Used:
The sine of an angle \[\theta \] in a right angled triangle is given by \[\sin \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Hypotenuse}}}}\].

Complete step-by-step answer:
First, we will draw the diagram using the information given in the question.

Here, AB and CD are the two banks of the river. The boat starts from the point R on the bank AB, making an angle of \[60^\circ \]. RQ is the distance travelled by the boat, that is 600 m.
We need to find the width of the river, that is PQ.
We will use the formula for sine of an angle of a right angled triangle to find the width of the river.
We know that the sine of an angle \[\theta \] in a right angled triangle is given by \[\sin \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Hypotenuse}}}}\].
In the triangle PQR, PQ is the perpendicular and QR is the hypotenuse.
Therefore, in triangle PQR, we have
\[ \Rightarrow \sin \angle PRQ = \dfrac{{PQ}}{{QR}}\]
Substituting \[\angle PRQ = 60^\circ \] and \[QR = 600\] m in the equation, we get
\[ \Rightarrow \sin 60^\circ = \dfrac{{PQ}}{{600}}\]
The sine of the angle measuring \[60^\circ \] is equal to \[\dfrac{{\sqrt 3 }}{2}\].
Substituting \[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\] in the expression, we get
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{{PQ}}{{600}}\]
Multiplying both sides by 600, we get
\[ \Rightarrow \dfrac{{600\sqrt 3 }}{2} = PQ\]
Thus, we get
\[ \Rightarrow PQ = 300\sqrt 3 \]
Therefore, the width of the river is \[300\sqrt 3 \] m.
We can find the approximate value of the width of the river by substituting \[\sqrt 3 = 1.732\].
Substituting \[\sqrt 3 = 1.732\] in the equation \[PQ = 300\sqrt 3 \], we get
\[ \Rightarrow PQ = 300 \times 1.732\]
Multiplying the terms, we get
\[ \Rightarrow PQ = 519.6\] m
Therefore, the width of the river is approximately \[519.6\] m.

Note: We used sine to solve the problem instead of tangent or cosine, because sine is the ratio of the perpendicular and the hypotenuse. The hypotenuse is the distance travelled by the boat, which is given. The perpendicular is the required width of the river. Therefore, using the sine helps to solve the problem much easily and in lesser steps than by using tangent or cosine.