Question

A circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30degree.

Answer 1

Answer:

Step-by-step explanation:

Let AB be the vertical pole and CA be the rope. Then,

∠ACB=30  

o

 and AC=20 m

In right △ ABC,

sin30  

o

=  

AC

AB

​  

 

2

1

​  

=  

20

AB

​  

 

AB=10 m

Therefore, the height of the pole is 10 m

Answer 2

Step-by-step explanation:

$\mathfrak{ \huge{ \green{ \underline{given}}}} \\ \mathfrak{ \large{ \red{ac \: = \: 20 \: m}}} \\ \mathfrak{ \large{ \red{angle \: of \: elevation = {30}^{o}}}} \\ \mathfrak{ \huge{ \green{ \underline{to \: find}}}} \\ \mathfrak{ \large{ \red{height \: of \: pole \: (h) \: = \: ?}}} \\ \mathfrak{ \huge{ \green{ \underline{formula \: to \: be \: used}}}} \\ \mathfrak{ \large{ \red{sin \: θ \: = \: \frac{perpendicular}{hypotenuse} }}} \\ \mathfrak{ \large{ \red{ \sin \: {30}^{o} \: = \: \frac{1}{2} }}} \\ \mathfrak{ \huge{ \green{ \underline{solution}}}} \\ \mathfrak{ \large{ \blue{sin \: θ \: = \: \frac{perpendiular}{hypotenuse} }}} \\ \mathfrak{ \large{ \blue{ \sin \: c \: = \: \frac{ab}{ac} }}} \\ \mathfrak{ \large{ \blue{ \sin \: {30}^{o} \: = \: \frac{ab}{20} }}} \\ \mathfrak{ \large{ \blue{ \frac{1}{2} \: = \: \frac{ab}{20} }}} \\ \mathfrak{ \large{ \blue{ab \: = \: \frac{20}{2} }}} \\ \mathfrak{ \large{ \blue{ab \: = \: 10}}} \\ \mathfrak{ \large{ \orange{ \underline{ => height \: of \: pole \: = \: 10 \: m}}}}$