Question

1. A circus artist is climbing a 20 m 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
30° (see Fig. 9.11). answe​

Answer 1

Answer:

the height of the rope will be 10 m

Step-by-step explanation:

length of rope is 20m

sin 30° = p/h = AB/AC

1/2 = AB/20

20 = 2 AB

20/2 = AB

10 = AB

Answer 2

Step-by-step explanation:

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