**Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2 Pdf**

TN Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2 – Trigonometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 6 Exercise 6.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 6.2

**TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry**

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
6 (Exercise 6.2) |

Chapter Name |
Trigonometry |

**TNSCERT Class 10th Maths Pdf | all Exercise Solution**

__Exercise – 6.2__

**(1) Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10****√3 m. **

**Solution: **

Let AB be the tower and the angle elevation from c is θ.

Given that AB = 10√3 m and BC = 30m

Now, ∆ABC

AB/BC = tanθ

tanθ = 10√3/30

tanθ = √3/3

tanθ = 1/√3

tanθ = tan30°

θ = 30°

Therefore, the required angles is 30°.

** **

**(2) A road is flanked on either side by continuous rows of houses of height 4√3 m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30°. Find the width of the road.**

**Solution: **

Let DB = x

Given that DC = BC = x/2

∴ BC = x/2

Given that, AB = 4√3

Now, ∆ABC,

tanθ = AB/BC

tan30° = 4√3/(x/2) [∵θ = 30°]

1/√3 = 4√3/(x/2)

x/2 = 4×3 = 12

x = 24

∴ BD = width of the road therefore 24m width of the road.

** **

**(3) ****To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (√3 = 1.732) **

**Solution: **

Now, CD = window

Let, CD = hm and AC = xm

Now, ∆ABC,

tanθ = AC/AB

tan45° = x/5 [∵θ = 45°]

1 = x/5

x = 5

∴ Now, AD = AC + CD = x + CD = (5 + h) m

Now, ∆ABD

tanθ = AD/AB

tan60° = 5+h/5 [∵θ = 60°]

√3 = 5+h/5

5+h = 5√3

h = 5√3 – 5

h = 5×1.732 – 5 [∵ √3 = 1.732]

h = 8.660 – 5

h = 3.66m

Thus, the window height is 3.66m.

** **

**(4) A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40°. Find the height of the pedestal. (tan40° = 0.8391, √3 = 1.732) **

**Solution: **

CD = statue and AC = pedestal.

Given that CD = 1.6m.

Let AC = h m and AB = xm

AD = AC + CD = (h + 1.6) m

∆ABC,

tan 40° = AC/AB

0.8391 = h/x [∵ tan40° = 0.8391]

x = h/0.8391 —- (i)

Now, ∆ABD,

tan60° = AD/AB

√3 = h+1.6/x

x = h + 1.6/√3

h/0.8391 = h+1.6/1.732 [By equation (i) and √3 = 1.732]

1.732h = 0.8391h + 1.6× 0.8391

1.732h – 0.8391h = 1.6 × 0.8391

0.8929h = 1.6 × 0.8391

h = 1.6×0.8391/0.8929

h = 1.6×8391/8929

h = 1.6×0.93

h = 1.5m

Thus, the pedestal height is 1.5m.

** **

**(5) A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. **

**Find **

**(i) the height of the pole **

**(ii) Radius of the dome. (****√3 = 1.732) **

**Solution: **

Now, AF = AD = radius = r

Now, AB = AF + FB = (r + 7) m.

Now, AC = AF + FB + BC = r + 7 + 5 = (r + 12) m.

and AE = AD + DE = (r + h) m.

∆ABE

tan45° = AE/AB

1 = r + h/r + 7

r + 7 = r + h

h = 7m

∆ACD,

tan30° = AD/AC

1/√3 = r/r+12

√3r = r + 12

(√3 – 1) r = 12

r = 12/√3 – 1

r = 12/1.732 – 1

r = 12/.732

r = 16.39m

Therefore, the height of the pole is 7m. and radius of the dome is 16.39 m.

** **

**(6) The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?**

**Solution: **

Given that AB = 15m

Let, BE = elective pole height = h

Let, BC = x m

AE = AB – BE = (15 – h) m

Now, ∆ABC,

tan60° = AB/BC

√3 = 15/x

x = 15/√3

x = 5√3 m

Now, ∆AED, tan30° = AE/ED

1/√3 = 15-h/5√3

15 – h = 5

h = 15 – 5 = 10m

Thus, the height of the elective pole is 40m.

*Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Exercise 6.2 *

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