In ΔEFG , what is the approximate length of EF if the measure of angle G is 50° and FG is 10 cm?

Mathematics

1 answer:

swat323 years ago

6 0

Hey again, you can use trigonometric ratios for this one. You are trying to find the side length opposite of angle G, when you have the adjacent side length value known. You can use TOA, or tan((opposite/adjacent)) . Tan(50) = x/10

X= 11.92 cm

You might be interested in

Help will give brainiest

son4ous [18]

Answer:

I'm pretty sure the answer is a

4 0

3 years ago

Read 2 more answers

Find first derivative f(x)= x² - 4 /x³ + 9

vitfil [10]

Answer:

f'(x) = 2x + $\frac{12}{x^{4} }$

Step-by-step explanation:

differentiate using the power rule

$\frac{d}{dx}$ ( a $x^{n}$ ) = na $x^{n-1}$

Given

f(x) = x² - $\frac{4}{x^3}$ + 9 = x² - 4 $x^{-3}$ + 9 , then

f'(x) = 2x + 12 $x^{-4}$ = 2x + $\frac{12}{x^{4} }$

5 0

2 years ago

Read 2 more answers

A guy wire to a tower makes a 72 degrees angle with level ground. At a point 30 ft farther from the tower than the wire but on t

Andreyy89

So.. checking the picture below

we can say y = y, or just do a substitution and end up with

$\bf tan(72^o)=tan(30^o)(30+x)\implies 3.08x=\cfrac{30}{\sqrt{3}}+\cfrac{1}{\sqrt{3}}x
\\\\\\
3.08x-\cfrac{1}{\sqrt{3}}x=\cfrac{30}{\sqrt{3}}\implies 2.5x\approx 17.32\implies x\approx \cfrac{17.32}{2.5}$

once you know, how long is "x", then you can simply use the cosine of 72° to get "r"

thus $\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(72^o)=\cfrac{x}{r}\implies r=\cfrac{x}{cos(72^o)}$