Question

this question number 10

Answer 1

Given : ABC is a field in the form of an Equilateral triangle

★  We know that : In an Equilateral triangle  ⇄  All Sides are Equal

Let the Length of the Side of the Field be : S

As : The Field is in the form of an Equilateral triangle

$:\implies$  AB = BC = CA = S

Given : A Vertical Pole of height 45 m is erected at A

Let the Top of the Vertical Pole erected at A be denoted by : E

Given : A Vertical Pole of height 20 m is erected at B

Let the Top of the Vertical Pole erected at B be denoted by : F

Now, Consider the Front View of the Field as if we are standing before the Boundary of the Field AB

Given : There is a point D on AB such that from it, the Angles of Elevation of the tops of the two poles are equal.

Let the Angle of Elevation of the two poles from Point D be : α

From Figure - 1 : Consider Tangent of Triangle AED

$\bigstar\;\;\boxed{\mathsf{Tangent\;of\;a\;Triangle = \dfrac{Opposite\;Side\;of\;the\;Triangle}{Adjacent\;Side\;of\;the\;Triangle}}}$

Here : Opposite side of the Triangle is AE

          Adjacent side of the Triangle is AD

$\mathsf{\implies Tan\alpha = \dfrac{AE}{AD}}$

$\mathsf{\implies Tan\alpha = \dfrac{45}{AD}}$

From Figure - 1 : Consider Tangent of Triangle DFB

Here : Opposite side of the Triangle is FB

          Adjacent side of the Triangle is DB

$\mathsf{\implies Tan\alpha = \dfrac{FB}{DB}}$

☯  DB can be written as : AB - AD

$\mathsf{\implies Tan\alpha = \dfrac{20}{AB - AD}}$

$\mathsf{\implies Tan\alpha = \dfrac{20}{S - AD}}$

☯  As both Tanα are Equal, We can equate them

$\mathsf{\implies \dfrac{45}{AD} = \dfrac{20}{S - AD}\;-----\;[1]}$

Consider the view standing at point C and watching the top of the tower at point A (i.e Point E)

Let the Angle of Elevation of this View be : β

From Figure - 2 : Consider Triangle ECA

Here : Opposite side of the Triangle is EA

          Adjacent side of the Triangle is CA

$\mathsf{\implies Tan\beta = \dfrac{EA}{CA}}$

$\mathsf{\implies Tan\beta = \dfrac{45}{S}}$

Consider the view standing at point C and watching the top of the tower at point B (i.e Point F)

Let the Angle of Elevation of this View be : Δ

From Figure - 3 : Consider Triangle CFB

Here : Opposite side of the Triangle is FB

          Adjacent side of the Triangle is CB

$\mathsf{\implies Tan\Delta = \dfrac{FB}{CB}}$

$\mathsf{\implies Tan\Delta = \dfrac{20}{S}}$

Given : Angles of Elevation of the tops of the two poles from C are Complementary to each other

$\mathsf{\bigstar\;\;Complementary\implies Sum\;of\;the\;Angles\;should\;be\;equal\;to\;90^{\circ}}$

$\mathsf{\implies \beta + \Delta = 90^{\circ}}$

$\mathsf{\implies \Delta = 90^{\circ} - \beta}$

$\mathsf{\implies Tan(90^{\circ} - \beta}) = \dfrac{20}{S}}$

★  We know that : Tan(90 - θ) = Cotθ

$\mathsf{\implies Cot\beta = \dfrac{20}{S}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{Cot\theta = \dfrac{1}{Tan\theta}}}}$

$\mathsf{\implies \dfrac{1}{tan\beta} = \dfrac{20}{S}}$

$\mathsf{\implies tan\beta = \dfrac{S}{20}}$

☯  As both Tanβ are Equal, We can equate them

$\mathsf{\implies \dfrac{45}{S} = \dfrac{S}{20}}$

$\mathsf{\implies S^2 = (45 \times 20)}$

$\mathsf{\implies S^2 = 900}$

$\mathsf{\implies S = \sqrt{900}}$

$\mathsf{\implies S = \sqrt{(\pm30)^2}}$

$\mathsf{\implies S = \pm30}$

$\mathsf{\implies S = 30\;\;(or)\;-30}$

$\mathsf{As\;Length\;cannot\;be\;Negative \implies S \neq -30}$

$\mathsf{\implies S = 30}$

$\implies \textsf{Length of the Side of the Equilateral Field = 30\;m}$

Substituting the value of S in Equation [1], We get :

$\mathsf{\implies \dfrac{45}{AD} = \dfrac{20}{30 - AD}}$

$\mathsf{\implies 45(30 - AD) = 20AD}$

$\mathsf{\implies 20AD = 45(30) - 45AD}$

$\mathsf{\implies 20AD + 45AD = 1350}$

$\mathsf{\implies 65AD = 1350}$

$\mathsf{\implies AD = \dfrac{1350}{65}}$

$\mathsf{\implies AD = \dfrac{270}{13}}$

$\mathsf{\implies AD = 20 + \dfrac{10}{13}}$

$\mathsf{\implies AD = 20\dfrac{10}{13}\;m}$