Question

Two trees \[A\] and \[B\] are on the same side of a river. From a point \[C\] in the river, the distances of the trees \[A\] and \[B\] are 250 m and 300 m respectively. If the angle \[C\] is \[45^\circ \], find the distance between the trees. \[\left( {{\rm{Use }}\sqrt 2 = 1.414} \right)\]

Answer 1

Hint:
Here, we need to find the distance between the trees \[A\] and \[B\]. First we draw the diagram based on the given information. We will use the law of cosines and the given information to find the distance between the trees \[A\] and \[B\].
Formula Used: We will use the formula for the law of cosines states that \[{c^2} = {a^2} + {b^2} - 2ab\cos C\], where \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle, and \[C\] is the angle opposite to the side of length \[c\].

Complete step by step solution:
First, we will draw the diagram as per the given information.

Here, \[C\] is the point in the river. \[BC\] is the distance between the tree \[B\] and point \[C\], and \[AC\] is the distance between the tree \[A\] and the point \[C\].
The distance between the trees \[A\] and \[B\] is \[AB\].
We will use the law of cosines to find the distance between the trees \[A\] and \[B\].
According to the law of cosine, \[{c^2} = {a^2} + {b^2} - 2ab\cos C\], where \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle, and \[C\] is the angle opposite to the side of the length \[c\].
Substituting \[a = BC\], \[b = AC\], and \[c = AB\] in the law of cosines, we get
\[ \Rightarrow A{B^2} = B{C^2} + A{C^2} - 2\left( {BC} \right)\left( {AC} \right)\cos C\]
Substituting \[BC = 300\]m, \[AC = 250\]m, and \[\angle C = 45^\circ \] in the equation, we get
\[ \Rightarrow A{B^2} = {\left( {300} \right)^2} + {\left( {250} \right)^2} - 2\left( {300} \right)\left( {250} \right)\cos 45^\circ \]
Applying the exponents on the bases, we get
\[ \Rightarrow A{B^2} = 90000 + 62500 - 2\left( {300} \right)\left( {250} \right)\cos 45^\circ \]
We know that the cosine of \[45^\circ \] is \[\dfrac{1}{{\sqrt 2 }}\].
Substituting \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] in the equation, we get
\[ \Rightarrow A{B^2} = 90000 + 62500 - 2\left( {300} \right)\left( {250} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow A{B^2} = 90000 + 62500 - \dfrac{{150000}}{{\sqrt 2 }}\]
Substituting \[\sqrt 2 = 1.414\] and simplifying, we get
\[\begin{array}{l} \Rightarrow A{B^2} = 90000 + 62500 - \dfrac{{150000}}{{1.414}}\\ \Rightarrow A{B^2} \approx 90000 + 62500 - 106082.03678\end{array}\]
Adding the terms in the expression, we get
\[ \Rightarrow A{B^2} \approx 46417.96322\]
Taking the square roots of both sides, we get
\[ \Rightarrow AB \approx \sqrt {46417.96322} \approx 215.45\]

\[\therefore \] We get the distance between the trees \[A\] and \[B\] as \[215.45\]m.

Note:
We used the law of cosines to solve the question. In any triangle \[ABC\], we can apply three laws of cosines.
(a) \[{a^2} = {b^2} + {c^2} - 2bc\cos A\], where \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle, and \[A\] is the angle opposite to the side of length \[a\].
(b) \[{b^2} = {a^2} + {c^2} - 2ac\cos B\], where \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle, and \[B\] is the angle opposite to the side of length \[b\].
(c) \[{c^2} = {a^2} + {b^2} - 2ab\cos C\], where \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle, and \[C\] is the angle opposite to the side of length \[c\].