Answer:
The answer is 12 units.
Step-by-step explanation:
Looking at the figure we see that
and
Now from equations (1) and (2) we have values for and
so we put those into the equation (3) and get:
Therefore the answer is 12 units.
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1 answer:
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The linear equations are y - 25 = 0.89(x - 20) and y = 0.89x + 7.2
<h3>The slope of the line</h3>The complete question is added as an attachment
The two points from the graph are (20, 25) and (38, 41)
The slope of the line is calculated using
m = (y2 - y1)/(x2 - x1)
Substitute the known values in the above equation
m = (41 - 25)/(38 - 20)
Evaluate
m =0.89
<h3>The linear equation in point slope form</h3>This is calculated as:
y - y1 = m(x - x1)
Substitute the known values in the above equation
y - 25 = 0.89 * (x - 20)
Evaluate
y - 25 = 0.89(x - 20)
<h3>The linear equation in slope-intercept form</h3>We have:
y - 25 = 0.89(x - 20)
Expand
y - 25 = 0.89x - 17.8
Add 25 to both sides
y = 0.89x + 7.2
Hence, the linear equations are y - 25 = 0.89(x - 20) and y = 0.89x + 7.2
Read more about linear equations at:
#SPJ1
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
