A student walks 50m on a bearing of 25 and then 200m due east. how far is she from her starting point
Your response to your own question has no resemblance at all
to anything in your problem. No idea what it is supposed to do or say.
I made a sketch and have triangle with sides 50 and 200 with an angle
of 115° between those sides, so, clearly I will use the cosine law.
d^2 = 50^2 + 200^2 - 2(50)(200)cos115°
= 50,952.365..
d = appr 225.7 m
Or
we could use vectors:
vector d = 50(cos65,sin65) + 200(cos0,sin0)
= (21.131,45.3154) + (200,0)
= (221.131, 45.3154)
|d| = √(221.131^2 + 45.3154^2)
= appr 225.7 , just like before
To find the distance from the starting point to the final point, we can use the concept of vector addition. This involves breaking down the distances into their horizontal and vertical components, and then adding those components together.
Let's break down the given distances:
Distance walked on a bearing of 25: 50m
Distance due east: 200m
Since the bearing of 25 is not specified as measured from magnetic north or true north, we'll assume it is the angle measured from the positive x-axis (east direction). Therefore, the horizontal component of the first distance can be found by multiplying the distance by the cosine of 25 degrees:
Horizontal component of the first distance = 50m * cos(25°)
Next, we can find the vertical component of the first distance by multiplying the distance by the sine of 25 degrees:
Vertical component of the first distance = 50m * sin(25°)
Thus, the horizontal component of the first distance is approximately 45.31m (rounded to two decimal places), and the vertical component is approximately 21.54m (rounded to two decimal places).
Now, let's add the horizontal components together:
Total horizontal component = horizontal component of the first distance + horizontal component of the second distance
Total horizontal component = 45.31m + 200m (since the second distance is due east)
Total horizontal component = 245.31m
And, let's add the vertical components together:
Total vertical component = vertical component of the first distance + vertical component of the second distance
Total vertical component = 21.54m + 0m (since the second distance is due east)
Total vertical component = 21.54m
Finally, we can find the distance from the starting point to the final point using the Pythagorean theorem:
Distance = √[(Total horizontal component)^2 + (Total vertical component)^2]
Distance = √[(245.31m)^2 + (21.54m)^2]
Distance = √[60147.2961m^2 + 464.1816m^2]
Distance = √[60511.4777m^2]
Distance ≈ 246m (rounded to the nearest meter)
Therefore, the student is approximately 246 meters from her starting point.
To find the distance the student is from her starting point, we can use the concept of vectors.
First, let's break down the two distances given into their respective components.
The first distance the student walks on a bearing of 25 can be split into two parts:
- The horizontal component (east-west) is given by 50 * cos(25)
- The vertical component (north-south) is given by 50 * sin(25)
The second distance of 200m due east has a horizontal component of 200 (as it is purely east).
Now, let's find the total horizontal and vertical components by adding up the individual components:
Horizontal component = 50 * cos(25) + 200 = approximately 197.85
Vertical component = 50 * sin(25) + 0 = approximately 21.13
To find the distance the student is from her starting point, we can use the Pythagorean theorem:
Distance = √(Horizontal component^2 + Vertical component^2)
Distance = √(197.85^2 + 21.13^2)
Distance = √(39142.8225 + 446.5969)
Distance = √39589.4194
Distance = approximately 198.96 meters
Therefore, the student is approximately 198.96 meters away from her starting point.
=13 × 2+6×2-2(13)(6) cos99
=169+36+24.4038
b×b=229.4038
b=√229.4038
=15.14608
=15km