A ship is stationary at sea.
1 answer:
Answer:
The distance between the yacht from the ship is about 33.6 km.
Step-by-step explanation:
The diagram is attached below.
Point A is the shipe, Point B is the tugboat, and Point C is the yacht.
Since the tugboat is 28 km away, and the scale being used is 1 cm : 4 km, the tugboat is 7 cm away from the ship at a 210° bearing on the diagram.
Likewise, the yacht is 6 cm away from the tugboat at a 210° bearing.
To find the distance between the yacht to the ship, we need to determine the value of <em>x</em>.
To do so, we can use the Law of Cosines:
First, we need to determine ∠B.
A has a bearing of 210°. 180° is covered by QI and IV. So, the small angle in QIII is 30°.
By Alternate Interior Angles, the angle in QI (inside the triangle) of Point B must also be 30°.
And since Point B has a bearing of 310°, 270° is covered in total by QI, QIV, and QIII. So, 40° is left in QII.
Therefore, 50° is the measure of the angle in QIII within the triangle of Point B.
Thus, ∠B measures 50° + 30° = 80°.
So, ∠B measures 80°. We also know that <em>a</em> = 6 and <em>c</em> = 7. Substitute:
Simplify and take the square root of both sides. So:
Approximate using a calculator:
Since 1 cm : 4km, to find the distance in km, we simply need to multiply by four. So:
The distance between the yacht from the ship is about 33.6 km.
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Answer:
The space that the package would occupy is 45 cubic feet.
Step-by-step explanation:
The space that the package occupies is the volume of the cylinder. The volume of an object is the 3 dimensional capacity of the object.
Volume of a cylinder = h
where: r is the radius and h is the height of the cylinder.
Given that; h = 5 feet and r = 3 feet.
The required volume = ×
× 5
= × 9 × 5
= 45 cubic feet
The space that the package would occupy is 45 cubic feet.
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
