All categories

3 years ago

Can someone help me out with this.

Mathematics

1 answer:

pickupchik [31]3 years ago

3 0

The side MK measures 3.3 feet.

Step-by-step explanation:

Step 1:

In the triangle, the angle is 25°, the adjacent side has a length of x feet while the hypotenuse of the triangle measures 3.6 feet. The opposite side of the triangle is not given.

To determine the value of x, we determine the value of the cos of the given triangle.

To calculate the cos of an angle we divide the adjacent side by the hypotenuse of the same triangle.

$cos \theta = \frac{adjacentside}{hypotenuse} .$

Step 2:

The length of the adjacent side = x feet.

The length of the hypotenuse = 3.6 feet.

$cos \theta = \frac{adjacentside}{hypotenuse}, cos 25 = \frac{x}{3.6} , cos 25 = 0.9063.$

$x = 0.9063 (3.6) = 3.2626.$

So the adjacent side of the triangle measures 3.2626 feet. Rounding this off to the nearest tenth, we get that MK measures 3.3 feet.

You might be interested in

Find the coordinate of the missing enpoint if B is the midpoint of AC. A(-4,2), B(6,-1) C

Eddi Din [679]

Step-by-step explanation:

Hey there!!

Here,

B(6,-1) is a midpoint of two end points AC.

A(-4,2) is a one end point.

Let another point be B(x,y).

Now, Using midpoint forumula.

$(x,y) = (\frac{x1 + x2}{2} , \frac{y1 + y2}{2} )$

Put all values.

$(6, - 1) = ( \frac{ - 4 + x}{2} ,\frac{2 + y}{2} )$

As they are equal, equating with their corresponding elements we get,

$6 = \frac{ - 4 + x}{2}$

12 = -4 + x

x = 16.

Again,

$- 1 = \frac{2 + y}{2}$

-2 = 2+ y

y = -4

Therefore, the coordinates were B(16,-4).

Hope it helps...

4 0

4 years ago

What is the slope of the line that passes through the points (2, 3) and (8, 6)?

Flauer [41]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What. Is the answer for 13-m=2(-4+m). Plz explain and show work

a_sh-v [17]

13-m=2(-4+m)
13-m=-8+m
-m-m = -8-13
-2m = -21
M= -21 : -2
= 10.5
Sorry if I'm wrong ..

8 0

3 years ago

The volume of a sphere is 1928m. What is the surface area of the sphere to the nearest tenth?

jasenka [17]

Answer:

749.1m

Step-by-step explanation:

Hope this helped!

4 0

2 years ago

For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on

nata0808 [166]

An algebraic equation enables the expression of equality between variable expressions

$\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}$

The reason the above value is correct is given as follows:

The given parameters are;

The symbol for the area of a triangle ΔXYZ = [XYZ]

The side length of the given square ABCD = 1

The location of point E = Side $\overline{BC}$ on square ABCD

The location of point F = Side $\overline{CD}$ on square ABCD

∠EAF = 45°

The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)

Required:

To find the value of [AEF]

Solution:

The area of a triangle = (1/2) × Base length × Height

Let x = EC, represent the base length of ΔCEF, and let y = CF represent the height of triangle ΔCEF

We get;

The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2

The area of ΔCEF, [CEF] = 1/9 (given)

∴ x·y/2 = 1/9

ΔABE:

$\overline{BE}$ = BC - EC = 1 - x

The area of ΔABE, [ABE] = (1/2)×AB ×BE

AB = 1 = The length of the side of the square

The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2

ΔADF:

$\overline{DF}$ = CD - CF = 1 - y

The area of ΔADF, [ADF] = (1/2)×AD ×DF

AD = 1 = The length of the side of the square

The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2

The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]

[ABCD] = Area of the square = 1 × 1

$[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}$

From $\dfrac{x \cdot y}{2} = \dfrac{1}{9}$ , we have;

$x = \dfrac{2}{9 \cdot y}$ , which gives;

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] = (1/2) × $\overline{AE}$ × $\overline{FA}$ × sin(∠EAF)

$\overline{AE}$ = √((1 - x)² + 1), $\overline{FA}$ = √((1 - y)² + 1)

[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

$\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Squaring both sides and plugging in $x = \dfrac{2}{9 \cdot y}$ , gives;

$\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}$

Subtracting the right hand side from the equation from the left hand side gives;

$\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0$

36·y² - 40·y + 8 = 0

$y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}$

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}$

Plugging in $y = \dfrac{5 + \sqrt{7} }{9}$ and rationalizing surds gives;

$[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}$

Therefore;

$\underline{[AEF]= \dfrac{4}{9}}$

Learn more about the use of algebraic equations here:

brainly.com/question/13345893

6 0

3 years ago