Solve a ship leave port and travel 21km on a bearing of 032° and then 45km on a bearing 287 calculate the distance and bearing o
1 answer:
Answer:
Step-by-step explanation:
Let shall be considering the Cosine rule in this situation.
a^2 = b^2+c^2- 2bc.cosA
Where a = ?
b = 21km
c 45km
A = 131
a^2 = 21^2+45^2-2(21*45).cos131
a^2 = 441+2025-1890(-0.6560)
a^2 = 2466+1239.951
a^2 = 3705.951
a= 60.876km
Finding the bearing of the port from the ship.
We shall be considering the Sine rule
Sin A/a = Sin B/b
Where A = ?
a = 45km
B = 131°
b = 60.876km
Sin A/45 = sin131/60.876
SinA×60.876 = 45×sin131
SinA×60.876= 45×0.7547
SinA×60.876 = 33.961
SinA= 33.961/60.876
SinA = 0.5578
A = sin^-1 0.5578
A = 33.9°
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Answer:
84 possible paths
Step-by-step explanation:
Given
--- 3 blocks north
--- 6 blocks east
Required
Number of distinct path
To solve this question, we make use of the following formula
The above formula implies that;
On a single path, there is a total of m + n steps to get to a particular position, where each path is either in m direction or n direction.
In this case:
So, we have:
Apply combination formula
Expand the numerator
Expand the denominator
<em>Hence, there are 84 possible paths</em>
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