All categories

3 years ago

HELP ME ASAP.. MY TEACHER IS ASKING FOR THE ANSWER

Mathematics

1 answer:

Solnce55 [7]3 years ago

5 0

Answer:

a) 10.7 cm

b) 11.6 cm

c) 29.9 cm²

Step-by-step explanation:

a) sin(α)/a = sin(β)/b

sin(∠BCD)/BD = sin(DBC)/DC

sin(41°)/BD = sin(55°)/13.4

BD = 13.4*sin(41°)/sin(55°) = 10.73 cm ≈ 10.7 cm

b) From ΔBCD , ∠BDC = 180-(41+55) = 84°

∠ABD and ∠BDC are alternate interior angles, so they are congruent, and

∠ABD = ∠BDC = 84°

AD²= AB² + BD² - 2*AB*BD*cos (∠ABD) =

= 5.6² + 10.73² - 2*5.6*10.73*cos(84°) = 133.93 cm²

AD =√(133.93) ≈11.6 cm

c) Area(ADB) = (1/2)*AB*BD*sin(∠ABD)=(1/2)*5.6*10.73*sin(84°) ≈ 29.9 cm²

You might be interested in

Use inductive and deductive reasoning to solve the problem. Draw a diagram and solve the problem Freddie the frog is a at the bo

Wewaii [24]

Answer:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Step-by-step explanation:

4 0

2 years ago

Which property can be justified using the ratios in triangle XYZ ?

Alenkasestr [34]

The correct answer is B.

Determine and recall first the trigonometric functions:

Sin A = Opposite/Hypotenuse
Cos A = Adjacent/Hypotenuse
Tangent = Opposite/Adjacent

If for example we try to take an angle, for example the Y, then Y is opposite Y. The Hypotenuse will be the longer side which is going to denoted by X and the next side will be Adjacent which will be denoted by Z.

Use this test to answer:

Sin Y = Y/X

where:

Y = opposite
X = hypotenuse

Cos (90 - y) = cos (90 - 50) = cos 40

where Z = 40
we look for cos Z = Y/X

Therefore,

Sin Y = Cos (90 - Y)

6 0

2 years ago

Read 2 more answers

Please help me I need help

sdas [7]

Answer:

3.51

Step-by-step explanation:

i took 9.2 ÷ 4.14 to find how much they charged per oz. then i took 7.8 and multipled it by 45 cents per oz.

5 0

2 years ago

Read 2 more answers

What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

=3.374

6 0

1 year ago

Solve for x. 1 2 x + 3 2 (x + 1) − 1 4 = 5 A) 5 2 B) 15 4 C) 15 8 D) 17 8

Hitman42 [59]

Answer: x=-13/44 or -.2954

8 0

2 years ago

Read 2 more answers

HELP ME ASAP.. MY TEACHER IS ASKING FOR THE ANSWER ​

HELP ME ASAP.. MY TEACHER IS ASKING FOR THE ANSWER