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shutvik [7]
3 years ago
11

Please help me on this

Mathematics
2 answers:
natita [175]3 years ago
7 0

Answer:

The value of cos\theta would be Base/Hypotenuse = EF/DF

Step-by-step explanation:

In figure, we are given a triangle

There is a 90 degree angle in the triangle. Therefore, the triangle is a right triangle.

In a right triangle,

Sin\theta = \frac{Perpendicular}{Hypotenuse}

Cos\theta = \frac{Base}{Hypotenuse}

Tan\theta = \frac{Perpendicular}{Base}

In triangle, the side with right angle and given angle is called base, Therefore

Base = EF

The side opposite to the right angle is called the hypotenuse, so

Hypotenuse = DF

and

Perpendicular = DE

Therefore, the value of cos\theta would be Base/Hypotenuse = EF/DF

klio [65]3 years ago
4 0
The answer is B because the answer is not a because anything/anything=1, and it is obviously not 1°, so it has to be b, unless there are more options, because it should be DF/DE
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Evaluate a(b - c ^ 2) * if * a = 2/3, b = 3/4, c = 1/2 A: 1/65 B: 1/3 C: 1/4 D: 2/3
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If a+b+c=1,

a

2

+

b

2

+

c

2

=

2

,

a

3

+

b

3

+

c

3

=

3

then find the value of

a

4

+

b

4

+

c

4

=

?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

)

⇒

2

(

a

b

+

b

c

+

c

a

)

=

1

2

−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

)

+

3

a

b

c

=

3

⇒

(

1

×

(

2

−

(

−

1

2

)

+

3

a

b

c

)

)

=

3

⇒

(

2

+

1

2

)

+

3

a

b

c

=

3

⇒

3

a

b

c

=

3

−

5

2

=

1

2

⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

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