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3 years ago

How to use Cosine to calculate the side of a right triangle?

Mathematics

1 answer:

svetlana [45]3 years ago

8 0

Answer:

Cosine is Adj/Hyp

a simple trick i use when doing these kind of problems is first finding the sides. If you only know the Hypotenuse then you multiply that with the cos(x) or whatever angle it is. If you only know Adjacent then you divide the adjacent with the Cos(x) and you get the answer

Step-by-step explanation:

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Can some plz help me!

Ymorist [56]

Answer:

a=9+6

p=-30+20

z=15+(-12)

x=-10+(-7)

Step-by-step explanation:

5 0

3 years ago

A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6

Burka [1]

Answer:

Tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

Step-by-step explanation:

Given : In the figure attached, Length of girl EC = 4 ft

Length of street light AB = 25 ft

Girl is moving away from the light with a speed = 6 ft per sec.

To Find : Rate ( $\frac{dw}{dt}$ ) of the tip (D) of the girl's shadow (BD) moving away from th

light.

Solution : Let the distance of the girl from the street light is = x feet

Length of the shadow CD is = y feet

Therefore, $\frac{dx}{dt}=6$ feet per sec. [Given]

In the figure attached, ΔAFE and ΔADE are similar.

By the property of similar triangles,

$\frac{x}{21}=\frac{x+y}{25}$

25x = 21(x + y)

25x = 21x + 21y

25x - 21x = 21y

4x = 21y

y = $\frac{4x}{21}$

Now we take the derivative on both the sides,

$\frac{dy}{dt}=\frac{4}{21}\times \frac{dx}{dt}$

= $\frac{4}{21}\times 6$

= $\frac{8}{7}$

≈ 1.14 ft per sec.

Since w = x + y

Therefore, $\frac{dw}{dt}= \frac{dx}{dt}+\frac{dy}{dt}$

$\frac{dw}{dt}=6+1.14$

= 7.14 ft per sec.

Therefore, tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

3 0

3 years ago

Find the x-intercept and y-intercept of the line.<br> 6x + 3y = - 18<br> I NEED HELP ASP

ser-zykov [4K]

X-intercept = -3
y- intercept = -6

6 0

3 years ago

Evaluate help please!

UNO [17]

<h2><u>ABSOLUTE VALUE</u></h2>

The absolute value of a number is the distance from 0 to that number. The distance is positive, hence, the absolute value is always a positive number.

<h3>Exercise</h3>

Replace the value of x:

$| x - 8 |$