Question

The angles of elevation of the top of a tower from two
points at a distance of 9 meters and 16 meters from the
base of the tower and in the same straight line with it, are
55° and 35° respectively. The height of the tower is​

Answer 1

Answer:

• Height of Tower = 12 metres

Explanation:

The angles of Elevation of the top of a tower from two points at a distance of 9 metres and 16 metres from the base of the tower in the same straight line are given as 55° and 35°

Refer to the attached image for figure

Now, Let height of tower be h

and since,

tan θ = opposite side / adjacent side

therefore,

→ tan 35° = h / 16   ___equation (1)

and,

→ tan 55° = h / 9

→ tan ( 90° - 35° ) = h / 9

Since, tan (90 - θ) = cot θ, therefore

→ cot 35° = h / 9

→ 1/tan 35° = h / 9

→ tan 35° = 9 / h

using equation (1)

→ h / 16 = 9 / h

→ h² = 16 × 9

→ h² = 144

→ h = 12 metres

Therefore,

• Height of Tower is 12 metres.

Answer 2

Answer:

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.7,2){\sf{\large{x}}}\put(9,0.5){\sf{\large{9 m}}}\put(10,0.1){\sf{\large{16 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.4,1.2){\sf\large{55^{\circ} }}\put(10.5,1.2){\sf\large{35^{\circ} }}\put(10,0.4){\vector( -1,0){2}}\put(10,0.4){\vector(1,0){2}}\put(9,0.8){\vector( -1,0){1}}\put(9,0.8){\vector(1,0){1.5}}\end{picture}$

Let the Height of Tower be x.

$\dashrightarrow\sf \tan( \theta) =\dfrac{Perpendicular}{Base}\\\\\\\dashrightarrow\sf \tan(55) =\dfrac{x}{9}\qquad\quad-eq.(1)\\\\\\\\\dashrightarrow\sf \tan(\theta) =\dfrac{Perpendicular}{Base}\\\\\\\dashrightarrow\sf \tan(35) =\dfrac{x}{16}\\\\\\\dashrightarrow\sf \tan(90-55) =\dfrac{x}{16}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{\tan(90-\theta)=\cot(\theta)}}\\\\\dashrightarrow\sf \cot(55) =\dfrac{x}{16}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{ \dfrac{1}{\tan(\theta)}=\cot(\theta)}}\\\\\dashrightarrow\sf \dfrac{1}{\tan(55)} =\dfrac{x}{16}\\\\\\\dashrightarrow\sf \tan(55) =\dfrac{16}{x}\qquad\quad -eq.(2)$

$\rule{150}{1.2}$

$\underline{\bigstar\:\textsf{From eq.(1) and eq.(2) :}}$

$:\implies\sf \tan(55)=\tan(55) \\\\\\:\implies\sf \dfrac{x}{9}=\dfrac{16}{x}\\\\\\:\implies\sf x \times x = 9 \times 16\\\\\\:\implies\sf {x}^{2} = 9 \times 16\\\\\\:\implies\sf x = \sqrt{9 \times 16} \\\\\\:\implies\sf x =(3 \times 4) \:cm\\\\\\:\implies\underline{\boxed{\sf x = 12 \:cm}}$

⠀

$\therefore\:\underline{\textsf{The Height of tower is \textbf{12 cm}}}.$