Question

Arun's house is in the main market of Karol bagh Delhi. As shown in the above picture there was an electric pole near Arun's house. The pole was getting a band and was disturbing the traffic on this road. BSES was informed by the father of Arun and was requested to solve this issue. To solve the issue the pole was changed and to prevent further banding of the pole, a supporting wire was fixed as shown in the picture. The supporting wire was fixed 4 m away from the pole. The wire is making an angle of 60o with the ground.
1. What is the value of cos 60o?
2. What is the height of the pole?
3. What is the length of the wire?
4. What is the area of the △ABC?
5. What is the value of tan 60o?
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Answer 1

Do 6419×45312 KAUSHAL5 to be in your area of your FAMILY for

Answer 2

Given:

Length of wire away from pole = 4m

Angle made by wire with ground = 60°

To find:

Value of Cos 60°

Height of pole

Length of wire

Area of ΔABC

Value of Tan 60°

Solution:

Since the pole is a straight line, it is ideally thought to be perpendicular to the ground, thus making 90° with the ground. If side AB represents the pole, BC represents the ground, then AC is the side which represents the supporting wire fixed on ground, that is banded to the pole.

Cos 60 = Adjacent side/Hypotenuse

$\frac{1}{2}=\frac{BC}{AC}$

$\frac{1}{2}=\frac{4}{AC}$

$AC=8m$

Tan 60 = opposite side/adjacent side

$\sqrt{3}=\frac{AB}{BC}$

$\sqrt{3}=\frac{AB}{4}$

$AB=4\sqrt{3}$ m

Area of ΔABC

= $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AB

= $\frac{1}{2}$ × $4$ × $4\sqrt{3}$

= $8\sqrt{3}$ $m^{2}$

1. Value of Cos 60° = $\frac{1}{2}$

2. Height of pole, AB = $4\sqrt{3}$ m

3. Length of wire = AC = 8m

4. Area of ΔABC = $8\sqrt{3} m^{2}$

5. Value of tan 60° = $\sqrt{3}$