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Andru [333]
2 years ago
11

The angle \theta_1θ

Mathematics
1 answer:
uysha [10]2 years ago
8 0

I guess that we know that the angle θ is in the fourth quadrant, and we know that:

cos(θ) = 9/19

now we want to find the value of sin(θ).

To do it, we can remember that for a point (x, y), such that we can define an angle β between the positive x-axis and a ray that connects the origin with the point (x, y), we can write the relations:

tan(β) = x/y

sin(β) = y/√(x^2 + y^2)

cos(β) = x/√(x^2 + y^2)

Because the angle is in the fourth quadrant, we know that:

x > 0

y < 0.

And we also know that:

cos(θ) = 9/19

then we have:

x = 9

√(x^2 + y^2) = √(9^2 + y^2) = √(81 + y^2) = 19

Solving the above equation we can find the value of y, that we need to remember, is negative:

√(81 + y^2) = 19

81 + y^2 = 19^2

y^2 = 19^2 - 81 = 280

y = √280 = -16.7

Now that we know the value of y, we can replace that in the sine equation to get:

sin(θ) = -16.7/19 = -0.879

If you want to learn more, you can read:

brainly.com/question/19830127

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Yochanan walked from home to the bus stop at an average speed of 5 km / h. He immediately got on his school bus and traveled at
kari74 [83]

Answer: 5 km walking and 30 km by bus

Step-by-step explanation:

Yochanan walked from home to the bus stop at an average speed of 5 km / h. He immediately got on his school bus and traveled at an average speed of 60 km / h until he got to school. The total distance from his home to school is 35 km, and the entire trip took 1.5 hours. How many km did Yochanan cover by walking and how many did he cover by travelling on the bus?

walking - 5km/h                                         bus - 60km/h

distance walking - d₁                                 distance bus - d₂

time walking - t₁                                         time bus - t₂

d₁ + d₂ = 35

t₁ + t₂ = 1.5

v = d/t

vwalking = d₁/t₁

5 = d₁/t₁ ⇒ d₁ = 5t₁

vbus = d₂/t₂

60 = d₂/t₂ ⇒ d₂ = 60t₂

d₁ + d₂ = 35 ⇒ 5t₁ + 60t₂ = 35

_________________________

5t₁ + 60t₂ = 35

t₁ + t₂ = 1.5  (*-5)

5t₁ + 60t₂ = 35

-5t₁ -5t₂ = -7.5   (+)

__________________________

55t₂ = 27.5

t₂ = 27.5/55 = 0.5 h

t₁ + t₂ = 1.5 ⇒ t₁ = 1.5 - 0.5 = 1h

d₁ = 5t₁ ⇒ d₁ = 5.1 = 5 km

d₂ = 60t₂ ⇒ d₂ = 30.0.5 = 30 km

6 0
2 years ago
This was given to me during a summative test and the teacher didn't bother giving me the correction. I just cannot figure it out
Dmitry [639]
  • Base be y

ATQ

\\ \sf\longmapsto xy=6050\dots 1

\\ \sf\longmapsto 2(x+y)=220\implies x+y=110\dots 2

Now

\\ \sf\longmapsto (x+y)^2=x^2+y^2+2xy

\\ \sf\longmapsto 110^2-2(6050)=x^2+y^2

\\ \sf\longmapsto 12100-12100=x^2+y^2

\\ \sf\longmapsto x^2+y^2=0\dots(3)

From all equations

\\ \sf\longmapsto (x-y)^2=x^2+y^2-2xy

\\ \sf\longmapsto (x-y)^2=0-2(6050)

\\ \sf\longmapsto (x-y)^2=-12100

\\ \sf\longmapsto (x-y)=110\dots(4)

Now

Adding 3 and 4

\\ \sf\longmapsto 2x=220

\\ \sf\longmapsto x=110m

One side won't be covered hence

  • y=2(110)=220m
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A set of the same elements from the set A and the set C.

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