All categories

3 years ago

In ΔDEF, the measure of ∠F=90°, DE = 8.8 feet, and EF = 6.4 feet. Find the measure of ∠D to the nearest tenth of a degree.

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

46.7 degrees.

Step-by-step explanation:

According to the given information, triangle DEF is a right triangle with angle F as the right angle. We are trying to find the measure of angle D.

We are given the opposite side length (6.4 feet) and the hypotenuse (8.8 feet), so we can use sine to calculate the angle measure (SOH = Sine; Opposite over Hypotenuse).

sin(D) = 6.4 / 8.8

sin(D) = 64 / 88

sin(D) = 8 / 11

D = arcsin(8 / 11)

D = arcsin(0.72727272727272727272727273)

D = 46.65824177.

So, the measure of angle D is about 46.7 degrees.

Hope this helps!

You might be interested in

The function f(x)={1/5}^x is translated up 4 units. Which equation represents the translated function

insens350 [35]

$f(x+n)\to\text{translation n units left}\\\\f(x-n)\to\text{translation n units right}\\\\f(x)+n\to\text{translation n units up}\\\\f(x)-n\to\text{translation n units down}\\\\\\f(x)=\left(\dfrac{1}{5}\right)^x\to f(x)+4=\left(\dfrac{1}{5}\right)^x+4\\\\Answer:\ g(x)=\left(\dfrac{1}{5}\right)^x+4$

3 0

3 years ago

Can some figure this out

NeX [460]

Hello,

As 4²+(5/2)²=22.25=89/4 (Pytagorean theorem)
Perimeter of the triangle =2*√89 /2 +5 =5+√89
Lateral area: 18*(5+√89)
Total area= 90+18√89+ 4*5/2 *2 =110+18√89 ≈279.811..(ft²)

8 0

4 years ago

Distribute (x-6)(x-7) a) x^2-13x-15 b) x^2-13x+42 c) x^2+x+42 d) x^2+x-15

Sidana [21]

Answer:

B. x² - 13x + 42

Step-by-step explanation:

7 0

3 years ago

How do you solve this equation : 5x-3y=2

monitta

By figuring out was is X and multiply

4 0

4 years ago

Read 2 more answers

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago