Question

A bridge across a river makes an angle of \[{{45}^{\circ }}\] with the river bank as shown in the figure. If the length of the bridge across the river is 150 m, what is the width of the river?

Answer 1

Hint: In this question we have to find the width of the river when we are given the length of the bridge and the angle made by the bridge with the bank of the river. It is given in the figure that there is formation of a right angle triangle in which the bridge is acting as the hypotenuse and the width o f the river is the perpendicular of the right angle triangle. As it is right angle triangle we can apply the trigonometric function like taking the sine of the given angle we can find the width of the river when length of the bridge is given, the relation becomes \[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}=\dfrac{x}{150}\].

Complete step-by-step answer:
In this question we are given that there is formation of a right angle triangle such that it's hypotenuse is the bridge laid over the river and the perpendicular of the triangle is the width of the river. We are also given that the angle made by the bridge with one of the bank of the river is \[{{45}^{\circ }}\] which means that the length of base of the triangle is equal to the length of the perpendicular of the triangle.

Now in the question we have to find the width of the river by using the information given above but first we need a relation to connect the width of the river with the length of the bridge and the angle made by the bridge with the banks of the river included in the triangle.
It is given that the triangle is right angle triangle so which means that we can apply the trigonometric functions on the triangle to find the missing elements of the triangle by using the values of known elements. In this case we need to find the width of the river which acts as the perpendicular to the triangle by using the length of bridge which acts as hypotenuse of the same, so we will use the sine function of the given angle which is as follows,

\[\sin \theta =\dfrac{Perpendicular}{Hypotenuse}\]
Let us assume the length of the perpendicular be \[x\].
Putting the known values in the function, we get,

\[\sin {{45}^{\circ }}=\dfrac{x}{150}\]
We need to remember the values of trigonometric functions on some particular angles in this case the value of \[\sin {{45}^{\circ }}\] function is \[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\].
Now solving the sine function when applied to the triangle, we get,
\[\dfrac{1}{\sqrt{2}}=\dfrac{x}{150}\]
Now the width of the river comes out to be \[x=\dfrac{150}{\sqrt{2}}\] or \[x=75\sqrt{2}\].
Note: One should remember the values of all 6 trigonometric functions at some particular angles. We can also do the question by using cosecant function \[\left( \operatorname{cosec}\theta =\dfrac{Hypotenuse}{Perpendicular}=\dfrac{1}{\sin \theta } \right)\] instead of using the sine function because both the functions include hypotenuse and perpendicular in their ratio formula. A short cut method to solve the question is by using the Pythagoras theorem which is applicable only in this case because both the angles of the triangle are \[{{45}^{\circ }}\] which means that the length of base and perpendicular are equal and when the theorem is applied we end up with only single variable equation which can be solved.