Mathematics, 05.01.2021 05:00 cupcakesmc619

find the area of each triangle. round intermediate values to the nearest tenth. use the rounded values to calculate the next value. Round your final answer to the nearest 10

find the area of each triangle. round intermediate values to the nearest tenth. use the rounded val

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Answer from: ryrytkg5107

Sin(α)=opposite leg/hypotenuse

Sin(63°)=ya/35

ya=35xSin(63°)

ya=31.2

Area(a)= (15.9x31.2)/2

Area(a)=248.0

Now, we proceed to find the area of "b". We already have the length x=15.9, so:

Tan(α)=opposite leg/adjacent leg

yb=15.90xTan(42°)

yb=14.3

Area(b)=15.9X14.3/2

Area(b)=113.7

The total Area (At) is:

At=Area(a)+Area(b)

At=248.0+113.7

At=361.7

10) Triangle "a":

-Adjacent leg of the triangle "a":

Cos(α)=adjacent leg/hypotenuse

Cos(30°)=xa/48

xa=48xCos(30°)

xa=41.6

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(30°)=y/48

y=48xSin(30°)

y=24.0

-Area of the triangle "a":

Area(a)=41.6x24.0/2

Area(a)=499.2

Triangle "b":

-We have the value of i(y=24).

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(45°)=24/xb

xb=24/Tan(45°)

xb=24

-Area of the triangle "b":

Area(b)= 24x24/2

Area(b)=288

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=499.2+288

At=787.2

11) To find the area of the triangle shown in this exercise, we have to apply the same procedure as in exercise 10:

-Adjacent leg of the triangle "a":

Cos(α) = adjacent leg/hypotenuse

Cos(56°)=xa/14

xa=14xCos(56°)

xa=7.8

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(56°)=y/14

y=14xSin(56°)

y=11.6

-Area of the triangle "a":

Area(a)=7.8x11.6/2

Area(a)=45.2

Triangle "b":

-The value of its opposite leg is y=11.6.

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(46°)=11.6/xb

xb=11.6/Tan(46°)

xb=11.2

-Area of the triangle "b":

Area(b)= 11.6x11.2/2

Area(b)=65.0

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=45.2+65.0

At=110.2

12)-Adjacent leg of the triangle "a":

Cos(α)=adjacent leg/hypotenuse

Cos(54°)=xa/13

xa=13xCos(54°)

xa=7.6

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(54°)=y/13

y=13xSin(54°)

y=10.5

-Area of the triangle "a":

Area(a)=7.6x10.5/2

Area(a)=39.9

Triangle "b":

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(42°)=7.6/xb

xb=7.6/Tan(42°)

xb=8.4

-Area of the triangle "b":

Area(b)=7.6x8.4/2

Area(b)=31.9

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=39.9+31.9

At=71.8First, it is important to remember that the formula to calculate the area of a triangle is: A=bxh/2

As we can see in the exercises, all the triangles are divided in two triangles. So, let's call "a" to the triangle on the left and "b" to the triangle on the right.

9)To find the area of the triangle "a", we need the lenght of the adjacent leg (x) and the opposite leg (ya):

Cos(α)=adjacent leg/hypotenuse

Cos(63°)=x/35

x=35xCos(63°)

x=15.9

Sin(α)=opposite leg/hypotenuse

Sin(63°)=ya/35

ya=35xSin(63°)

ya=31.2

Area(a)= (15.9x31.2)/2

Area(a)=248.0

Now, we proceed to find the area of "b". We already have the length x=15.9, so:

Tan(α)=opposite leg/adjacent leg

yb=15.90xTan(42°)

yb=14.3

Area(b)=15.9X14.3/2

Area(b)=113.7

The total Area (At) is:

At=Area(a)+Area(b)

At=248.0+113.7

At=361.7

10) Triangle "a":

-Adjacent leg of the triangle "a":

Cos(α)=adjacent leg/hypotenuse

Cos(30°)=xa/48

xa=48xCos(30°)

xa=41.6

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(30°)=y/48

y=48xSin(30°)

y=24.0

-Area of the triangle "a":

Area(a)=41.6x24.0/2

Area(a)=499.2

Triangle "b":

-We have the value of i(y=24).

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(45°)=24/xb

xb=24/Tan(45°)

xb=24

-Area of the triangle "b":

Area(b)= 24x24/2

Area(b)=288

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=499.2+288

At=787.2

11) To find the area of the triangle shown in this exercise, we have to apply the same procedure as in exercise 10:

-Adjacent leg of the triangle "a":

Cos(α) = adjacent leg/hypotenuse

Cos(56°)=xa/14

xa=14xCos(56°)

xa=7.8

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(56°)=y/14

y=14xSin(56°)

y=11.6

-Area of the triangle "a":

Area(a)=7.8x11.6/2

Area(a)=45.2

Triangle "b":

-The value of its opposite leg is y=11.6.

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(46°)=11.6/xb

xb=11.6/Tan(46°)

xb=11.2

-Area of the triangle "b":

Area(b)= 11.6x11.2/2

Area(b)=65.0

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=45.2+65.0

At=110.2

12)-Adjacent leg of the triangle "a":

Cos(α)=adjacent leg/hypotenuse

Cos(54°)=xa/13

xa=13xCos(54°)

xa=7.6

-Opposite leg of the triangle "a":

Sin(α)=opposite leg/hypotenuse

Sin(54°)=y/13

y=13xSin(54°)

y=10.5

-Area of the triangle "a":

Area(a)=7.6x10.5/2

Area(a)=39.9

Triangle "b":

-Adjacent leg of the triangle "b":

Tan(α)=opposite leg/adjacent leg

Tan(42°)=7.6/xb

xb=7.6/Tan(42°)

xb=8.4

-Area of the triangle "b":

Area(b)=7.6x8.4/2

Area(b)=31.9

-Total area of the triangle (At) is:

At=Area(a)+Area(b)

At=39.9+31.9

At=71.8

Step-by-step explanation:

Answer from: Quest

answer: $\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

$\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

which is the required equation for finding the value of x,

$\frac{x}{sin40^{\circ}}= \frac{12}{sin60^{\circ}}$

$x= 12\times \frac{sin40^{\circ}}{sin60^{\circ}}$

$x= 12\times 0.74222719896$

$x= 8.90672638762\approx 8.967$

Answer from: Quest

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