Answer:
Part A
The bearing of the point 'R' from 'S' is 225°
Part B
The bearing from R to Q is approximately 293.2°
Step-by-step explanation:
The location of the point 'Q' = 35 km due East of P
The location of the point 'S' = 15 km due West of P
The location of the 'R' = 15 km due south of 'P'
Part A
To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that
=
, the right triangle ΔRPS is an isosceles right triangle
∴ ∠PRS = ∠PSR = 45°
The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°
Part B
∠PRQ = arctan(35/15) ≈ 66.8°
Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°
Question:
You are in a bike race. When you get to the first checkpoint, you are 2/5 of the distance to the second checkpoint. When you get to the second check point, you are 1/4 of the distance to the finish. If the entire race is 40 miles, what is the distance between the start and the first check point?
Answer: 4 miles
Step-by-step explanation:
Let distance between start to first checkpoint = x
First checkpoint to second checkpoint = 2/5 of x
Distance of start to checkpoint 1 = ( 2/5 of start to checkpoint 2)
Distance of start to checkpoint 2 = (1/4 of start to finish)
If start to checkpoint 2 = 1/4 of start to finish
Then,
Distance of start to checkpoint 1 = ( 2/5 * 1/4 of start to finish)
Distance of start to checkpoint 1 = 2/20 of start to finish = 1/10 of start to finish
Entire race = 40 miles = distance from start to finish
1/ 10 of 40
= ( 1/10) × 40
= 4 miles

Step-by-step explanation:
The ratio cosine tell us that
- In a right angled triangle, the cosine of an angle is the side adjacent to the angle divided by the hypotenuse of the triangle.
In other words,

The side adjacent to angle a is 9. The hypotenuse is 41.
So

Answer:

Step-by-step explanation:
<u>Graphs</u>
The graph shows the amount of water in gallons as a function of time in seconds.
We are required to find the amount of water in the bucked after 1 second. Since the graph does not show an exact location for t=1, we must find the equation of the line.
Two points are clear on the graph: (0,0) (12,5). This gives us the necessary information to find the equation of the line, which has the form:

Where w is the amount of water in the bucket, t is the time, and m and b are two constants to be determined.
Using the point (0,0):

It follows that b=0
The equation now takes the form:

Using the point (12,5):

We find m:

The final equation is:

Substituting t=1:

The bucket has 5/12 gallons of water at t=1 second. Expressing the answer as an ordered pair:

Answer:
Area = 40mm
Example:
The below-solved example problem may be useful to understand how the values are being used in the mathematical formulas to find the trapezoid area.
Example Problem :
Find the area of a trapezoid having the bottom length a = 15 cm, top length b = 12 & height h = 9 cm?
Solution :
The given values
bottom length a = 15 cm
top length b = 12 cm
height h = 10 cm
Step by step calculation
formula to find area = ((a + b)/2) h
substitute the values
= ((15 + 12)/2) x 9
= 121.5 cm2