Question

The shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sun rays meet the ground at \[60^\circ \]. Find the angle between the sun rays and the ground at the time of the longer shadow.
(a) \[45^\circ \]
(b) \[30^\circ \]
 (c) \[15^\circ \]
(d) \[90^\circ \]

Answer 1

Hint: Here, we will use the formula for tangent in the triangle having the shorter shadow as the base, then write the height in terms of the length of the shorter shadow. Then, using this and the tangent in the triangle having the longer shadow as the base, we will find the angle between the sun rays and the ground at the time of the longer shadow.

Formula Used:
We will use the formula of the tangent of an angle \[\theta \] in a right angled triangle which is given by \[\tan \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\].

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

Here, AB is the flagstaff. BC is the length of the shadow when the sun rays meet the ground at \[60^\circ \]. BD is the length of the longer shadow.
Let the height of the flagstaff be \[h\].
Let the length of the shadow when the sun rays meet the ground at \[60^\circ \] be \[x\].
Thus, we get
\[ \Rightarrow BC = x\]
Now, it is given that the shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sun rays meet the ground at \[60^\circ \].
Therefore, using the figure, we get
\[ \Rightarrow BD = 3BC\]
Substituting \[BC = x\] in the expression, we get
\[ \Rightarrow BD = 3x\]
Now, we will use the formula for tangent of an angle of a right angled triangle to find the angle between the sun rays and the ground at the time of the longer shadow.
We know that \[\tan \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\].
Therefore, in triangle \[ABC\], we have
\[ \Rightarrow \tan \angle BCA = \dfrac{{AB}}{{BC}}\]
Substituting \[\angle BCA = 60^\circ \], \[BC = x\] and \[AB = h\] in the equation, we get
\[ \Rightarrow \tan 60^\circ = \dfrac{h}{x}\]
The tangent of the angle measuring \[60^\circ \] is equal to \[\sqrt 3 \].
Substituting \[\tan 60^\circ = \sqrt 3 \] in the expression, we get
\[ \Rightarrow \sqrt 3 = \dfrac{h}{x}\]
Multiplying both sides of the equation by \[x\], we get
\[ \Rightarrow \sqrt 3 \times x = \dfrac{h}{x} \times x\]
Thus, we get
\[ \Rightarrow h = \sqrt 3 x\]
Now, using the formula for tangent in triangle \[ABD\], we have
\[ \Rightarrow \tan \angle BDA = \dfrac{{AB}}{{BD}}\]
Substituting \[AB = h\] and \[BD = 3x\] in the equation, we get
\[ \Rightarrow \tan \angle BDA = \dfrac{h}{{3x}}\]
Substituting \[h = \sqrt 3 x\], we get
\[ \Rightarrow \tan \angle BDA = \dfrac{{\sqrt 3 x}}{{3x}}\]
Simplifying the expression, we get
\[ \Rightarrow \tan \angle BDA = \dfrac{{\sqrt 3 }}{3}\]
Rewriting 3 as the product of \[\sqrt 3 \] and \[\sqrt 3 \], we get
\[ \Rightarrow \tan \angle BDA = \dfrac{{\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}\]
Thus, we get
\[ \Rightarrow \tan \angle BDA = \dfrac{1}{{\sqrt 3 }}\]
Now, we know that the tangent of the angle measuring \[30^\circ \] is equal to \[\dfrac{1}{{\sqrt 3 }}\].
Substituting \[\dfrac{1}{{\sqrt 3 }} = \tan 30^\circ \] in the expression, we get
\[ \Rightarrow \tan \angle BDA = \tan 30^\circ \]
Therefore, we get
\[ \Rightarrow \angle BDA = 30^\circ \]
\[\therefore \] The angle between the sun rays and the ground at the time of the longer shadow is \[30^\circ \].
Thus, the correct option is option (b).

Note: We used tangent to solve the problem instead of sine or cosine, because tangent is the ratio of the perpendicular and the base. The base is the length of the shadow. The perpendicular AB is common to both triangles ABC and ABD. Therefore, using the tangent helps to solve the problem much more easily than sine or cosine.