MATHEMATICS HIGH SCHOOL

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠B1 is larger than ∠B2.) a = 31, c = 42, ∠A = 39° ∠B1 = ° ∠B2 = ° ∠C1 = ° ∠C2 = ° b1 = b2 =

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

In triangle ABC we have $a = 31, c = 42, ∠A = 39°$

To find other sides and angles.

Use sine formula for triangles

$(a)/(sin A) =(b)/(sin B) =(c)/(sinC) \n(31)/(sin39) =(b)/(sinB) =(42)/(sinc)$

Cross multiply to get

$Angle C =58.5^(o)$ or $121.5$

Angle B = 180-A-C

= $82.5^(o)$ or $19.5$

b =18.84 or 16.44

There are two triangles

with

$Side a = 31\nSide b = 48.84\nSide c = 42\nAngle ∠A = 39° \nAngle ∠B = 82.50° \nAngle ∠C = 58.5°$

$Side a = 31\nSide b = 16.44204\nSide c = 42\nAngle ∠A = 39° \nAngle ∠B = 19.5° \nAngle ∠C = 121.5°$

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What is the difference between class limits and class boundaries?

Answers

Answer:

Step-by-step explanation:

Class limits are the minimum data value(lower) and maximum data value (upper) that a class can contain. They usually have the same numerical accuracy as the original data values.

Class boundaries are boundary lines that mark or separate where one class stops and the other begins. The lower class boundary of a given class is got by finding the average of the previous upper class limit and the given lower class limit while the upper class boundary is got by finding the average of the given upper class limit and the next lower class limit.

Final answer:

Class limits and class boundaries are statistical terms used in frequency distributions. Class limits are the smallest and largest values of a class, while class boundaries are the points separating one class from another.

Explanation:

The terms class limits and class boundaries are used in the field of statistics, particularly in the context of frequency distributions. Class limits are the smallest and largest values that can fall within each class in a frequency distribution, whereas class boundaries are the points that separate one class from another, and each boundary forms the end of one class and the start of the next.

For example, imagine you are analyzing the frequency of test scores and you have a class with limits of 80 and 89. These limits are the smallest and largest scores that fit into that class. However, the class boundaries are 79.5 and 89.5, serving as the dividing lines between this class and the next ones.

Learn more about Class Limits and Class Boundaries here:

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Suppose that the equation of motion for a particle (where s is in meters and t in seconds) is s = 5 t^2 - 8 t + 3
find the acceleration at the instant when the velocity is at 0

Answers

Differentiate equation
v = 10t - 8
Differerentiate again
a = 10

Accelleration will always be 10m/s^-2

Ten to the fourth power as a product of tens

Answers

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Karla is riding vertically in a hot air balloon, directly over a point P on the ground. Karla spots a parked car on the ground at an angle of depression of 30o. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?

Answers

Answer:

407 m

Step-by-step explanation:

Refer the attached figure

Karla spots a parked car on the ground at an angle of depression of 30° i.e. ∠BDP = 30°

Now the balloon rises 50 meters.i.e. AB = 50 m

So, the angle of depression to the car is 35 degrees. i.e.∠ADP = 35°

Let DP be the distance of the car prom point P

Let BP be x

In ΔBDP

$Tan\theta = (Perpendicular)/(Base)$

$Tan 30^(\circ) = (BP)/(DP)$

$(1)/(√(3))= (x)/(DP)$

$DP= (x)/((1)/(√(3)))$ --1

In ΔADP

$Tan\theta = (Perpendicular)/(Base)$

$Tan 35^(\circ) = (AP)/(DP)$

$0.70020= (50+x)/(DP)$

$DP= (50+x)/(0.70020)$ --2

So, equating 1 and 2

$(50+x)/(0.70020) = (x)/((1)/(√(3)))$

$(50+x)/(x) = (0.70020)/((1)/(√(3)))$

$(50+x)/(x) =1.21278$

$50+x =1.21278x$

$50 =1.21278x-x$

$50 =0.21278x$

$(50)/(0.21278)=x$

$x=234.982$

Substitute the value of x in 1

$DP= (234.982)/((1)/(√(3)))$

$DP= 407.0007$

Hence the car is at a distance of approximately 407 m from point P

We form 2 similar rt. triangles:

Triangle #1.
X = Hor. side = dist. from P to the car.
Yi = Ver. side.
A = 30o.

Triangle #2.
X = Hor. side = Dist. from P to the car.
Y2 = Y1+50 = Ver. side.
A = 35o.

Use aw of Sines.
sin30/Y1 = sin35/(Y1+50).
Cross multiply:
(Y1+50)sin30 = Y1*sin35
Y1/2+25 = 0.574Y1
Multiply both sides by 2:
Y1+50 = 1.148Y1
1.148Y1-Y1 = 50
0.148Y1 = 50
Y1 = 337.8 m.

tan30 = Y1/X = 337.8/X
X = 337.8/tan30 = 585 m. 714 feet away hope that helps if not tell me.

Cost A=0.6489

take the inverse cosine of both sides

Answers

I am not quite sure what the question asks for,
But this is what i assume it wants:

Cos A= 0.6489
In this given one, we basically find the size of the angle A
we do cosine inverse on both sides to get the size of the angle A
$cos^(-1)$ : It looks like this in the calculator
$cos^(-1)$ × cos A= $cos^(-1)$ (0.6489)
( $cos^(-1)$ and cos cancels out)
A= $cos^(-1)$ (0.6489)
A=49.54°
check:
cos 49.54=0.6489 (its right!)

One side of an isosceles triangle has a length of 19 m. The lengths of the other two sides are equal to one another, but are unknown. If the perimeter of the triangle is 51 m, what is the length of each unknown side?

Answers

Let x= side 1= 19
Let y= side 2=unknown
Let z= side 3=unknown
Since it is an isosceles triangle, two sides are equal, which happen to be y and z
Hence y=z

Perimeter of a triangle = Sum of all three sides
= x+y+z
51=19+2y [since we know x=19 and y=z, we know the perimeter is 51]
Solve now
32=2y
16=y
Since y=z
z= 16
Hence the unknown side's length is 16m

So the perimeter is 51. The base is 19. Subtract 19 from 51. 51 - 19 = 32. So 32 is the total length of the two equal unknown sides added together. Because the two sides must be equal, you can just divide 32 by 2. 32/2 = 16. Each of the two congruent sides are 16 m.

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