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3 years ago

A car travels 32 km due north and

Mathematics

1 answer:

musickatia [10]3 years ago

5 0

Answer:

Ifa,bandcare the

sides opposite to anglesA,B,&C.

By Cosine Rule,

=32

+46

−(2×32×46cos140°)

=1024+2116−2944cos140°

=3140−2944cos140°

=3140+2255.234841

=5395.234841

b=73.45

By Sine Rule,

sinC

sin140°

73.45

sinC=

73.45

32sin140°

sinC=0.2800

C=16.26°

The direction of the resultant

vector is measured from0°east

to the resultant, this angle isθ−90°,

90°is the total angle in

the first quadrant.

∴θ−90°=C+ϕ

ϕ=50°{Alternate angles are equal}

∴The direction of the car’s

resultant vector is50°+16.26°=66.26°south of east

Step-by-step explanation:

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k0ka [10]

Answer:

$2489ft^{2}$

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

$A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}$

<h3>Area of each circular section:</h3>

From the figure, we know that $\alpha =2.21 radians$ and the radius is $R=30ft$ . So, the are would be calculated with this formula:

$A_{cs}=\frac{\pi R^{2}\alpha}{360\°}$

Replacing all values:

$A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}$

Remember that $360\°=6.28radians$

Therefore, $A_{cs}=994.5ft^{2}$

Now, the total are of the figure is:

$A_{total}=2A_{t}+2A{cs}=2(250ft^{2} )+2(994.5ft^{2})\\A_{total}=500ft^{2} + 1989ft^{2}=2489ft^{2}$

Therefore the area of the symmetrical pool is $2489ft^{2}$

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Step-by-step explanation: