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Yanka [14]
3 years ago
10

this is due in less the 30 minutes and this is my last question on the assignment please help and show work. will give brainlies

t

Mathematics
2 answers:
Oduvanchick [21]3 years ago
5 0

Answer:

22,100

Step-by-step explanation:

valina [46]3 years ago
5 0

Answer:

the answer in 18,300

Step-by-step explanation:

2 x 3 5/8= 7 1/4

7 1/4 +1 9/10= 9 3/20

9 3/20 / 2= 18 3/10

18 3/10 x 1000=18,300

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2 years ago
Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>​
Ksivusya [100]

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

 sin  B

=

c

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

h

c

    a n d        sin  C =

h

b

or

h = c  sin  B     a n d       h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

      a n d          sin  C =

h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

 sin  C

7 0
3 years ago
A man is standing at a radar base and observes an unidentified plane at an altitude 6000m flying towards the radar base at an an
liubo4ka [24]

Answer:

The speed in of the plane is 115.47 m/sec

Step-by-step explanation:

Given:

Height at which the plane is flying = 6000 m

Angle of elevation at the radar base = 30 Degrees

Angle of elevation at the radar base after one minute  = 60 Degrees

To Find:

The Speed of the plane in meter per second = ?

Solution:

Let us use the tangent of the angle to find the distance (d) to a point directly below plane:

<u>when the angle is 30 degrees</u>

tan(30) = \frac{6000}{d1}

d1   = \frac{6000}{tan(30)}

d1 = \frac{6000}{0.577}

d1 = 10392.3 meters

<u>when the angle is 60 degrees</u>

tan(60) = \frac{6000}{d2}

d2  = \frac{6000}{tan(60)}

d2  = \frac{6000}{1.732}\\

d2 = 3464.1 meters

<u>distance travelled by aircraft in 1 min is  </u>

=>d1 - d2

=>0392.3 - 3464.1

= 6928.2 m/min

<u>Now converting to m/sec</u>

=>\frac{6928.2}{60}

=>115.47 m/sec

4 0
3 years ago
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