Solve this fast plz.........!!!!!!

1.	Solve this fast plz.........!!!!!!
Answer» Diagram: $\setlength{\unitlength}{1cm} \begin{picture}(6,6)\thicklines \put(2, 2){\line(0,1){0.4}} \put(4, 2){\line(0,1){2.4}} \put(6, 2){\line(0,1){2.4}} \put(2,2){\line(1,0){4}} \put(2,2.4){\line(1,0){4}} \put(2,2.4){\line(1,1){2}} \put(4,4.4){\line(1,0){2}} \put(2,2){\line(1,0){4}} \put(2,2.4){\line(1,9){4}} \put(6,4.5){\sf B} \put(4,4.5){\sf A} \put(6.1,2.5){\sf C} \put(6.1,1.8){\sf D} \put(4,1.7){\sf E} \put(2,1.7){\sf F} \put(1.8,2.5){\sf G} \put(1,2.2){\sf 1.2m} \put(2.6,2.8){${60}^{ \circ} $}\put(2.6,2.5){${30}^{ \circ} $} \put(4.2,2){\vector(0,1){1.6}}\put(4.2,2){\vector(0, - 1){0}}\put(4.3,3){\sf 88.2m} \put(4.3,2.1){\sf 1.2m} \put(6,2.1){\sf 1.2m} \put(3.7,2.5){\sf H} \end{picture}$ Given: Height of Girl = 1.2m Balloon is flying at the height of 88.2m from the Ground. The ANGLE of elevation of the balloon from the EYES of girl at any INSTANT is 60° After some time angle of elevation reduced to 30° Find: Distance Travelled by Balloon HC Solution: Let, HC = x Now, In $\triangle$ AGH AH = AE - HE AH = 88.2 - 1.2 = 87m AH = 87m Now, $\dashrightarrow \sf \dfrac{B}{P} = \dfrac{HG}{AH} = \cot {60}^{ \circ} \\ \\$ where, AH = 87m cot 60° = 1/√3 So, $\dashrightarrow \sf \dfrac{HG}{AH} = \cot {60}^{ \circ} \\ \\$ $\dashrightarrow \sf \dfrac{HG}{87} = \dfrac{1}{ \sqrt{3}} \\ \\$ $\qquad\quad$ ☯ Doing Cross-multiplication☯ $\dashrightarrow \sf \dfrac{HG}{87} = \dfrac{1}{ \sqrt{3}} \\ \\$ $\dashrightarrow \sf HG \sqrt{3} = 87 \\ \\$ $\dashrightarrow \sf HG= \dfrac{87}{ \sqrt{3}} \\ \\$ $\qquad\quad$ ☯ Rationalising Denominator☯ $\dashrightarrow \sf HG= \dfrac{87}{ \sqrt{3}} \\ \\$ $\dashrightarrow \sf HG= \dfrac{87}{ \sqrt{3}} \times \dfrac{ \sqrt{3} }{ \sqrt{3} } \\ \\$ $\dashrightarrow \sf HG= \dfrac{87 \sqrt{3} }{3}m\\ \\$ $\therefore \sf HG= \dfrac{87 \sqrt{3} }{3}m$ •❅──────────✧❅✦❅✧──────────❅• Now, In $\triangle$ BGC $\dashrightarrow \sf \dfrac{B}{P} = \dfrac{GC}{BC} = \cot {30}^{ \circ} \\ \\$ where, GC = x + 87√3/3 BC = 87m cot 30° = √3 So, $\dashrightarrow \sf \dfrac{GC}{BC} = \cot {30}^{ \circ} \\ \\$ $\dashrightarrow \sf \dfrac{x + \dfrac{87 \sqrt{3} }{3} }{87} = \sqrt{3} \\ \\$ $\qquad\qquad$ ☯ Taking L.C.M☯ $\dashrightarrow \sf \dfrac{x + \dfrac{87 \sqrt{3} }{3} }{87} = \sqrt{3} \\ \\$ $\dashrightarrow \sf \dfrac{\dfrac{3x + 87 \sqrt{3} }{3} }{87} = \sqrt{3} \\ \\$ $\dashrightarrow \sf \dfrac{3x + 87 \sqrt{3} }{3} = 87 \times \sqrt{3} \\ \\$ $\dashrightarrow \sf \dfrac{3x + 87 \sqrt{3} }{3} = 87\sqrt{3} \\ \\$ $\dashrightarrow \sf 3x + 87 \sqrt{3} = 3(87\sqrt{3}) \\ \\$ $\dashrightarrow \sf 3x + 87 \sqrt{3} = 261\sqrt{3} \\ \\$ $\dashrightarrow \sf 3x = 261\sqrt{3} - 87 \sqrt{3} \\ \\$ $\dashrightarrow \sf 3x = 174\sqrt{3} \\ \\$ $\dashrightarrow \sf x = \dfrac{174\sqrt{3}}{3} \\ \\$ $\dashrightarrow \sf x = 58\sqrt{3}m \\ \\$ $\therefore \sf HC = x = 58\sqrt{3}m \\ \\$ _____________________ Hence, The Distance Travelled by Balloon is 58√3m

Answer»

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6)\thicklines \put(2, 2){\line(0,1){0.4}} \put(4, 2){\line(0,1){2.4}} \put(6, 2){\line(0,1){2.4}} \put(2,2){\line(1,0){4}} \put(2,2.4){\line(1,0){4}} \put(2,2.4){\line(1,1){2}} \put(4,4.4){\line(1,0){2}} \put(2,2){\line(1,0){4}} \put(2,2.4){\line(1,9){4}} \put(6,4.5){\sf B} \put(4,4.5){\sf A} \put(6.1,2.5){\sf C} \put(6.1,1.8){\sf D} \put(4,1.7){\sf E} \put(2,1.7){\sf F} \put(1.8,2.5){\sf G} \put(1,2.2){\sf 1.2m} \put(2.6,2.8){${60}^{ \circ} $}\put(2.6,2.5){${30}^{ \circ} $} \put(4.2,2){\vector(0,1){1.6}}\put(4.2,2){\vector(0, - 1){0}}\put(4.3,3){\sf 88.2m} \put(4.3,2.1){\sf 1.2m} \put(6,2.1){\sf 1.2m} \put(3.7,2.5){\sf H} \end{picture}$

Given:

Height of Girl = 1.2m
Balloon is flying at the height of 88.2m from the Ground.
The ANGLE of elevation of the balloon from the EYES of girl at any INSTANT is 60°
After some time angle of elevation reduced to 30°