Question

Solve this fast plz.........!!!!!!​

Answer 1

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 2

Answer:

$hii$

have a nice day ✌️✌️✌️✌️