need answers asap to this question given below
Answers
Step-by-step explanation:
Solutions :-
1)
From the given figure
Consider ∆ ABC ,
AB = 20 units
DC = 30 units
∠ABC = X°
∠CAD = 60°
∠ADC = ∠ ADB = 90°
In ∆ ADC,
tan 60° = Opposite side to 60° / Adjacent side to 60°
=> tan 60° = DC / AD
=> tan 60° = 30/AD
=> √3 = 30/AD
=> AD × √3 = 30
=> AD = 30/√3
=> AD = (10×√3×√3)/√3
=> AD = 10√3 units
Now,
In ∆ ADB,
sin x° = Opposite side to x° / Hypotenuse
=> sin x° = AD / AB
=> sin x° = 10√3/20
=> sin x° = √3/2
=> sin x° = sin 60°
Therefore, x ° = 60°
The value of x = 60°
2)
Given that
3 cos 80° cosec 10° + 2 sin 59° sec 31°
=> 3 cos(90°-10°) cosec 10°+2 sin (90°-31°) sec 31°
We know that
sin (90° - A) = cos A
cos (90°-A) = sin A
=> 3 sin 10° cosec 10° + 2 cos 31° sec 31°
=> 3 sin 10° (1/sin 10°) + 2 cos 31° (1/cos 31°)
=> 3 (sin 10°/sin 10°) + 2 (cos 31°/cos 31°)
=> 3 (1) + 2(1)
=> 3+2
=> 5
Therefore,
3cos 80°cosec 10°+2cos 59°sec 31° = 5
Used formulae:-
→ sin A = Opposite side to A / Hypotenuse
→ tan A = Opposite side to A / Adjacent side to A
→ sec A = 1/ cos A
→ cosec A = 1/ sin A
→ sin (90°- A) = cos A
→ cos (90° - A) = sin A
→ tan 60° = √3
→ sin 60° = √3/2
Step-by-step explanation:
Step-by-step explanation:
Solutions :-
1)
From the given figure
Consider ∆ ABC ,
AB = 20 units
DC = 30 units
∠ABC = X°
∠CAD = 60°
∠ADC = ∠ ADB = 90°
In ∆ ADC,
tan 60° = Opposite side to 60° / Adjacent side to 60°
=> tan 60° = DC / AD
=> tan 60° = 30/AD
=> √3 = 30/AD
=> AD × √3 = 30
=> AD = 30/√3
=> AD = (10×√3×√3)/√3
=> AD = 10√3 units
Now,
In ∆ ADB,
sin x° = Opposite side to x° / Hypotenuse
=> sin x° = AD / AB
=> sin x° = 10√3/20
=> sin x° = √3/2
=> sin x° = sin 60°
Therefore, x ° = 60°
The value of x = 60°
2)
Given that
3 cos 80° cosec 10° + 2 sin 59° sec 31°
=> 3 cos(90°-10°) cosec 10°+2 sin (90°-31°) sec 31°
We know that
sin (90° - A) = cos A
cos (90°-A) = sin A
=> 3 sin 10° cosec 10° + 2 cos 31° sec 31°
=> 3 sin 10° (1/sin 10°) + 2 cos 31° (1/cos 31°)
=> 3 (sin 10°/sin 10°) + 2 (cos 31°/cos 31°)
=> 3 (1) + 2(1)
=> 3+2
=> 5
Therefore,
3cos 80°cosec 10°+2cos 59°sec 31° = 5
Used formulae:-
→ sin A = Opposite side to A / Hypotenuse
→ tan A = Opposite side to A / Adjacent side to A
→ sec A = 1/ cos A
→ cosec A = 1/ sin A
→ sin (90°- A) = cos A
→ cos (90° - A) = sin A
→ tan 60° = √3
→ sin 60° = √3/2