Question

A ladder 5m long, leans against a wall so that it makes an angle of 60° with the horizontal ground. Calculate how far up the wall the ladder reaches.

Answer 1

Step-by-step explanation:

here AB is the wall.BC is the ground.AC is the length of ladder which is 5 meters. in triangle ABC, sin(60) degrees = AB/AC we know sin(60) degrees= (√3)/2. therefore (√ 3)/2 = AB/5. therefore AB=(5√3)/2

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.4,2){\sf{\large{? m}}}\put(9.3,2){\sf{\large{5 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{60^{\circ} }}\end{picture}$

$\rule{120}{0.8}$

Given : Let a Ladder AC be 5m long, makes angle of 60° with the horizontal ground.

To Find : How far up the wall the ladder reaches.

$\underline{\bigstar\:\sf{According\:to\:the\:Question :}}$

$\dashrightarrow\sf\:\:\sin(C)=\dfrac{Perpendicular}{Hypotenuse}\\\\\\\dashrightarrow\sf\:\:\sin(60^{\circ})=\dfrac{AB}{AC}\\\\\\\dashrightarrow\sf\:\: \dfrac{\sqrt{3}}{2} = \dfrac{BC}{5\:m}\\\\\\\dashrightarrow\sf\:\:BC = \dfrac{5\sqrt{3}\:m}{2}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf BC \approx 4.32\:m}}$

⠀

$\therefore\:\underline{\textsf{Ladder will reach to \textbf{4.32 m} approximate height on wall}}.$

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$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$