GSEB std 10 science solution for Gujarati check Subject Chapters Wise::
જવાબ : tanx = Height/Base tan30° = Height/10 Height= 10/√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 10
જવાબ : tanx = Height/Base Tan60° = Height/10 Height= 10√3m
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 10
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tan60° = 
Base = 5
જવાબ : tanx = Height/Base tan30° = Height/5 Height= 5/√3m
જવાબ : tanx = Height/Base Tan60° = Height/5 Height= 5√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 5
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base tan30° = Height/30 Height= 10√3m
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 30
જવાબ : tanx = Height/Base
tan30°= 
Base = 30
જવાબ : tanx = Height/Base Tan60° = Height/30 Height=30√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 1
જવાબ : tanx = Height/Base tan30° = Height/1 Height= 1/√3m
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx= 
= 1
x= 45°
જવાબ : tanx = Height/Base
tan30° = 
- Base = 40
જવાબ : tanx = Height/Base tan30° = Height/40 Height= 40/√3m
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 25
જવાબ : tanx = Height/Base Tan60° = Height/25 Height= 25√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 25
જવાબ : tanx = Height/Base tan30° = Height/25 Height= 25/√3m
જવાબ : tanx = Height/Base
tanx= 
= 1
x= 45°
જવાબ : tanx = Height/Base
Tan45° = 
Base = 10
જવાબ : tanx = Height/Base Tan45° = Height/10 Height= 10m
જવાબ : tanx = Height/Base
tan0° = 
Base = 0
જવાબ : tanx = Height/Base tan0° = Height/10 Height= 0m
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx=
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 20
જવાબ : tanx = Height/Base Tan60° = Height/20 Height= 20√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 20
જવાબ : tanx = Height/Base tan30° = Height/20 Height= 20/√3m
જવાબ : tanx = Height/Base
tanx=
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 100
જવાબ : tanx = Height/Base Tan60° = Height/100 Height= 100√3m
જવાબ : tanx = Height/Base
tan30° = 
Base = 100
જવાબ : tanx = Height/Base tan30° = Height/100 Height= 100/√3m
જવાબ : tanx = Height/Base
tanx= 
= √3
x= 60°
જવાબ : tanx = Height/Base
tanx= 
= 1/√3
x= 30°
જવાબ : tanx = Height/Base
tan60° = 
Base = 1
જવાબ : tanx = Height/Base Tan60° = Height/1 Height= √3m
જવાબ : 
∠BAC = 180° – 90° – 60° = 30°
sin 30° = BC/AC
1/2=BC/15
2BC = 15
BC = 15/2m
જવાબ : AB = 20 m, BC =20√3m,
In ∆ABC, Using trigonometry
જવાબ : BD = AB – AD = 6 – 2.54 = 3.46 m
In rt., ∆DBC, sin 60° = BD/DC
√3/2=3.46/DC
√3 /DC = 3.46 x 2
∴ Length of the ladder, DC = 6.92 /√3=6.92/1.73
DC = 4 m
જવાબ : Let us assume AC be the ladder
∴ Length of ladder, AC = 5 m 2.5 m
જવાબ : Let us assume required angle be θ.
tan θ = 30 /10 √3
tan θ = √3
⇒ tan θ = tan 60°
જવાબ : Let us assume that A be the point on the ground and C be the aeroplane.
In ∆ABC,
જવાબ : Let us assume that AC be the building & DE be the tower.
∴ Height of the tower, DE = BC = DE = AC – ABDE = 60 – 20[latex]\sqrt{3}[/latex] = 20(3 – [latex]\sqrt{3}[/latex])
DE = 20(3 – 1.73) = 20(1.27)
DE = 25.4 m
જવાબ : 
જવાબ : Let us assume that AL = x, BL = CM = 1500√3
જવાબ : 
BD = AE = 50 m
=> BC = CD – BD = (h – 50) m
From (i),
x = h – 50 = 118.25 – 50 = 68.25 mHeight of the tower, h = 118.25 m
∴ Horizontal distance between tower and Building, x = 68.25 m
જવાબ : 
Here BC is the distance of hill from ship and CE be the height of hill.
In rt. ∠ABC, tan 30° = AB/BC
BC = 10√3 m .(A)
BC = 10(1.73) = 17.3 m
AD = BC = 10√3m …(B) [From (A)
In rt. ∆ADE, tan 60° = DE/AD
⇒ √3 =DE/10 √3 … [From (B)
⇒ DE = 10√3 × √3 = 30 m
∴ CE = CD + DE = 10 + 30 = 40 m
Hence, the distance of the hill from the ship is 17.3 m and the height of the hill is 40 m.
જવાબ : Let the base of towers be k.
જવાબ : Let us assume that AB be the height of the light house,
D and C are two ships
- DC = 200 m
Let BC = x m, AB = h m
In rt. ∆ABC,
∴ Height of the light house = 273 m
જવાબ : Let us assume that AB be the 1st pole
and CE be the 2nd pole = 24 m
In ∆ADE,
=> Height of the first pole,
AB = CD = CE – DE
= 24 – 8.66 = 15.34 m
જવાબ : Let AB = 60 m be the height of Light-house and C and D be the two ships.
In right ∆ABC,
=> Distance between the two ships = CD = BD – BC = 103.92 – 60 = 43.92 m
જવાબ : 
Let us assume thzat AB be the light house and C and D be the two ships,
In ∆ABC, tan 45° = AB/BC
જવાબ : Letus assume OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and ∠ OAB = 90° and ∠AOB = 60°, let OB = l meters.
From the right triangle OAB, we have
જવાબ : 
જવાબ : 
જવાબ : 
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જવાબ :
1-C, 2-A, 3-B, 4-D
જવાબ :
1-B, 2-D, 3-A, 4-C
જવાબ :
1-B, 2-D, 3-A, 4-C
જવાબ :
1-B, 2-C, 3-D, 4-A
જવાબ :
1-A, 2-C, 3-B
જવાબ :
1-D, 2-C, 3-B, 4-A
જવાબ :
1-D, 2-C, 3-A, 4-B
જવાબ :
1-D, 2-C, 3-A, 4-B
જવાબ :
1-B, 2-C, 3-D, 4-A
જવાબ :
1-C, 2-A, 3-B, 4-D
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Some Applications of Trigonometry
Math
Chapter 09 : Some Applications of Trigonometry
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