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3 years ago

How to find the area of a right angle triangle with one side length given?

1 answer:

5 0

Step 1 Decide which two sides we are using - one we are trying to find and one we already know – out of Opposite, Adjacent and Hypotenuse.Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question.Step 3 For Sine write down Opposite/Hypotenuse, for Cosine write down Adjacent/Hypotenuse or for Tangent write down Opposite/Adjacent. One of the values is the unknown length.Step 4 Solve using your calculator and your skills with Algebra

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Please help soon! Will give branliest!

rosijanka [135]

Answer:

1. .26

2. .29

3. .22

4. .23

Step-by-step explanation:

3.1 ÷ 12 = .26

2.92 ÷ 10 = .29

3.52 ÷ 16 = .22

1.85 ÷ 8 = .23

<em>good luck, i hope this helps :)</em>

3 0

3 years ago

Read 2 more answers

Given: Triangle ABC, mmBC=24 cm, line segment AL- < bisector

Furkat [3]

The value of line AL is 21. 51cm

<h3>How to determine the length</h3>

To find line AL,

Using

Sin α = opposite/ hypotenuse to find line AB

Sin 90 = x/ 24

1 = x/24

Cross multiply

x = 24cm

Now, let's find line AC

Sin angle B = line AC/24

Note that to find angle B

angle A + angle B + angle C = 180

But angle B = 2 Angle A

x + 2x + 90 = 180

3x + 90 = 180

3x = 180-90

x = 30°

Angle B = 2 × 30 = 60°

Sin 60 = x/ 24

0. 8660 = x/24

Cross multiply

x = 24 × 0. 8660

x = 20. 78cm

We have the angle of A in the given triangle to be divide into two by the bisector, angle A = 15°

To find line AL, we use

Cos = adjacent/ line AL

Cos 15 = 20. 78/ line AL

Line AL = 20. 78/ cos 15

Line AL = 20. 78 / 0. 9659

Line AL = 21. 51 cm

Thus, the value of line AL is 21. 51cm

Learn more about trigonometry ratio here:

brainly.com/question/24349828

#SPJ1

6 0

2 years ago

A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that

rosijanka [135]

Using it's concept, it is found that the sample mean package weight for Thursday is of 19 ounces.

The mean of a data-set is given by the <u>sum of the observations divided by the number of observations</u>.

For Thursday, the package weights in ounces are given by: {18, 19, 20, 19}.

Hence, the sample mean in ounces is of:

M = (18 + 19 + 20 + 19)/4 = 19.

More can be learned about the mean concept at brainly.com/question/24628525

3 0

2 years ago

Can someone answer these as well worth lots of points. Only Appropriate Answers Pls :).

Rashid [163]

1. To solve this we are going to find the distance of the three sides of each triangle, and then, we are going to use Heron's formula.
- For triangle RTS:
S (2,1), T (1,3) and R (5,5). Using a graphic tool or the distance formula $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$ we realize that ST=2.2, SR=5, and TR=4.5. Now we are going to find the semi-perimeter of our triangle:
$s= \frac{ST+SR+TR}{2}$
$s= \frac{2.2+5+4.5}{2}$
$s= \frac{11.7}{2}$
$s=5.85$
Now we can use Heron's formula:
$A= \sqrt{s(s-ST)(s-SR)(s-TR)}$
$A= \sqrt{5.85(5.85-2.2)(5.85-5)(5.85-4.5)}$
$A=4.9$ square units
We can conclude that the area of triangle RTS is 4.9 square units.

- For triangle MNL:
M (2,-4), N (-2,-3), and L (0,-1). Once again, using a graphic tool or the distance formula we get that MN=4.1, ML=3.6, and NL=2.8.
Lets find the semi-perimeter of our triangle to use Heron's formula:
$s= \frac{4.1+3.6+2.8}{2}$
$s=5.25$
$A= \sqrt{5.25(5.25-4.1)(5.25-3.6)(5.25-2.8)}$
$A=4.9$ square units
We can conclude that the area of triangle MNL is 4.9 square units.

2. To find the volume of our pyramid, we are going to use the formula for the volume of a rectangular pyramid: $V= \frac{lwh}{3}$
where
$l$ is the length of the rectangular base
$w$ is the width of the rectangular base
$h$ is the height of the pyramid
From our picture we can infer that $l=15$ , $w=10$ , and $h=12$ , so lets replace the values in our formula:
$V= \frac{lwh}{3}$
$V= \frac{(15)(10)(12)}{3}$
$V=600$ cubic units

To find the volume of the cone, we are going to use the formula: $V= \pi r^2 \frac{h}{3}$
where
$r$ is the radius
$h$ is the height
From our picture we can infer that the diameter of the cone is 9; since radius is half the diameter, $r= \frac{9}{2} =4.5$ . We also know that $h=12$ , so lets replace the values:
$V= \pi r^2 \frac{h}{3}$
$V= \pi (4.5)^2 \frac{12}{3}$
$V=254.5$ cubic units
We can conclude that the volume of the pyramid is 600 cubic units and the volume of the cone is 254.5 cubic units

3. To find $y$ , we are going to use the tangent trigonometric function:
$tan( \alpha )= \frac{opposite.side}{adjacent.side}$
$tan(53)= \frac{4cm}{y}$
$y= \frac{4cm}{tan(53)}$
$y=3.01cm$
and to the nearest whole number:
$y=3cm$
We can conclude that $y=3cm$

4. To solve this, we are going to find the volume of the cube first; to do it we are going to use the formula for the volume of a cube: $V=a^3$
where
$a$ is the edge of the cube
Since the sphere fits exactly in the cube, the edge of the cube is equal to the diameter of the sphere; therefore, $a=12$ . Lets replace the value in our formula:
$V=(12)^3$
$V=1728$ cubic units
Next, we are going to use the formula for the volume of a sphere: $V= \frac{4}{3} \pi r^3$
where
$r$ is the radius of the sphere
We know form our problem that the diameter of the sphere is 12 units, since radius is half the diameter, $r= \frac{12}{2} =6$ . Lest replace the value in our formula:
$V= \frac{4}{3} \pi r^3$
$V= \frac{4}{3} \pi (6)^3$
$V=904.8$ cubic units
Now, to find the volume of the empty space in the cube, we just need to subtract the volume of the sphere form the volume of the cube:
$V_{empty}=1728-904.8=823.2$ cubic units
We can conclude that the volume of the empty space in the cube is 823.2 cubic units

4. To solve this, we are going to use the arc length formula: $s= \frac{ \alpha}{360} 2 \pi r$
where
$s$ =mBC
$s$ is the length of the arc
$r$ is the radius of the circle
Lets, find $x$ first. From the picture we can infer that:
$x+2x+x=180$
$4x=180$
$x= \frac{180}{4}$
$x=45$
Since BE is the diameter of the circle, AB is its radius; therefore $r=AB$ . Lets replace the values in our formula:
$s= \frac{ \alpha}{360} 2 \pi r$
$s= \frac{45}{360} 2 \pi AB$
$s= \frac{ \pi }{4} AB$
$mBC= \frac{ \pi }{4} AB$
We can conclude that $mBC= \frac{ \pi }{4} AB$

3 0

3 years ago

Please help I'll give brainliest! And 100 points this is due tonight!!!

AlladinOne [14]

Answer:

You can try searching it up or going to math.way

Step-by-step explanation:

8 0

3 years ago