Question

from a point a,level with the foot of a vertical pole and 25m from it ,the angle of elevation of the top of the pole is 48 degree .calculate
the ht of the pole
the angle of elevation from a of a point half way up the pole ​

Answer 1

$\bf{\underline{\green{Given:-}}}$

⚠ Distance between point A and foot of vertical pole = 25m

⚠ Angle of elevation of the top of the wall = 48°

$\bf{\underline{\green{Find:-}}}$

⚠ Height of the Pole

$\bf{\underline{\green{Diagram:-}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.3,0.3)}\put(.9,4.6){\large\bf P}\put(.8,.5){\large\bf Q}\put(5.5,.5){\large\bf A}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7) \qbezier(5.5,1)(1,3)(1,3)\put(3.8,1.2){ \large\bf 48^{\circ} } \put(.5,2.8){\large\bf R} \put(3,.6){\large\bf 25m} \end{picture}$

$\bf{\underline{\green{Solution:-}}}$

In ∆PAQ, right angled at Q

$\sf \Longrightarrow tan \: 48^{ \circ} = \dfrac{P}{B}$

$\sf \Longrightarrow tan \: 48^{ \circ} = \dfrac{PQ}{AQ}$

where,

• AQ = 25m
• tan 48° = 1.11061

So,

$\sf \dashrightarrow tan \: 48^{ \circ} = \dfrac{PQ}{AQ}$

$\sf \dashrightarrow 1.11061 = \dfrac{PQ}{25}$

$\bf \bigstar Cross - Multiplication \bigstar$

$\sf \dashrightarrow 1.11061 \times 25 = PQ$

$\sf \dashrightarrow 27.765m(approx.) = PQ$

$\sf \dashrightarrow 27.8m= PQ$

$\sf \dashrightarrow PQ = 27.8m$

$\sf \therefore PQ = 27.8m$

Thus, the height of the pole is 27.8m