Question

if a ladder of 15m long, inclined at the top of the vertical wall of height 15/2m, then find the angle of the ladder with ground​

Answer 1

Answer:

# = 30°

Step-by-step explanation:

consider as a right angled triangleABC ( let take teta as #)

BC = 15/2m , AC = 15m

here AC is the hypotenuse and BC is opposite

so , sin# = opp/hyp , = 15/2/15 , = 15/30

sin# = 1/2 , # = 30° , since sin30°=1/2

so the angle is 30°

Answer 2

Answer: 30°

Given:

$\setlength{\unitlength}{20} \begin{picture}(6,6) \linethickness{1.25}\put(1,1){\line(0,1){4}} \put(1,1){\line(1,0){3}} \qbezier(4,1)(4,1)(1,5) \put(1,0.5){ \bf B } \put(4,0.5){ \bf C }\put(1,5.25){ \bf A }\put(1.25,1){\line(0,1){0.25}}\put(1,1.25){\line(1,0){0.25}}\put(2.25,0.5){ \tt \frac{15}{2}\:m }\put(0.5,3){ \tt } \put(3,3){ \tt 15\:m } \end{picture}$

To find:

$\sf Angle\:of\:the\:ladder\:with\:ground.$

Explanation:

$\sf AC=15\:m (Length\:of\:the\:ladder)$

$\sf AB=\frac{15}{2}=7.5\:m(Height\:of\:the\:wall)$

$\sf Let,$

$\sf Angle\:of\:the\:ladder\:with\:the\:ground=\theta$

$\sf Now,$

$\sf We\:know\:that\:sin\theta\:is\:the\:ratio\:of\:the\:perpendicular\:to\:the\:hypotenuse.Therefore,$

$\sf In\: \Delta\:ABC,$

$\sf sin\theta=\dfrac{Perpendicular}{Hypotenuse}$

$\sf sin\theta=\dfrac{7.5}{15}$

$\sf sin\theta=0.5=\dfrac{1}{2}$

$\sf sin\:30^{\circ}=\dfrac{1}{2}$

$\sf Hence,$

$\boxed{\sf \theta=30^{\circ}}$

Therefore, the angle of the ladder with the ground​ is 30°.