Question

14. 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 figure)
A
20 m
30°
B
C​

Answer 1

Answer:

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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.

solution

Answer 2

Step-by-step explanation:

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