Question

plz send 2 questions
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Answer 1

Solution :-

2)

Given that

sin A = √3/2

=> sin A = sin 60°

Therefore, < A = 60°

Now,

cosec² A - 1

= cosec² 60° - 1

= (2/√3)²-1

= (4/3)-1

= (4-3)/3

= 1/3

and

sec² A + 1

= sec² 60° + 1

= 2²+1

= 4+1

= 5

Now,

(cosec² A +1)/(sec² A +1 )

= (1/3)/5

= 1/(3×5)

= 1/15

Therefore, (cosec² A +1)/(sec² A +1 ) = 1/15

3)

In ∆ DEF , < DEF = 90°

DE = EF

=> < EDF = < EFD

Since, The angles opposite to equal sides are equal.

< EDF = < EFD = 45°

DF = 20√2 cm

We know that

sin 45° = DE / DF

=> 1/√2 = DE / 20√2

=>√2 DE = 20√2

=> DE = 20√2/√2

=> DE = 20 cm

Therefore, DE = EF = 20 cm

We know that

The perimeter of a triangle is the sum of the all sides.

Perimeter of ∆ DEF = DE + EF + DF

=> 20+20+20√2

= 40+20√2 cm

= 40+20(1.414)

= 40+ 28.18

= 68.18 cm

= 68.2 cm

Perimeter of ∆ DEF = 40+20√2 cm or 68.2 cm

Answer 2

Step-by-step explanation:

Solution :-

2)

Given that

sin A = √3/2

=> sin A = sin 60°

Therefore, < A = 60°

Now,

cosec² A - 1

= cosec² 60° - 1

= (2/√3)²-1

= (4/3)-1

= (4-3)/3

= 1/3

and

sec² A + 1

= sec² 60° + 1

= 2²+1

= 4+1

= 5

Now,

(cosec² A +1)/(sec² A +1 )

= (1/3)/5

= 1/(3×5)

= 1/15

Therefore, (cosec² A +1)/(sec² A +1 ) = 1/15

3)

In ∆ DEF , < DEF = 90°

DE = EF

=> < EDF = < EFD

Since, The angles opposite to equal sides are equal.

< EDF = < EFD = 45°

DF = 20√2 cm

We know that

sin 45° = DE / DF

=> 1/√2 = DE / 20√2

=>√2 DE = 20√2

=> DE = 20√2/√2

=> DE = 20 cm

Therefore, DE = EF = 20 cm

We know that

The perimeter of a triangle is the sum of the all sides.

Perimeter of ∆ DEF = DE + EF + DF

=> 20+20+20√2

= 40+20√2 cm

= 40+20(1.414)

= 40+ 28.18

= 68.18 cm

= 68.2 cm

Perimeter of ∆ DEF = 40+20√2 cm or 68.2 cm