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pishuonlain [190]

3 years ago

6

A 18 foot ladder is leaning against a wall as shown

1 answer:

Inga [223]3 years ago

6 0

I'm guessing the diagram shows a ladder leaning against a wall, making a right angle triangle with respect to the ground and the wall.

So, the wall's height is going to be the 'h', which will also be the 'opposite side' from the angle <span>ϴ which is made from the ladder and the ground.
</span>The ladder's length (18 foot) is going to be the 'hypotenuse' side and the other remaining side will be the 'adjacent'.

Now, once you've sorted out which side is which, we have to find the h (opp), and according to SOH CAH TOA, we will choose Sin<span>ϴ = opp/hyp.
</span>so Sinϴ = h/18....now we gotta find h, so 'cross multiply' the equation to get h = 18 x sin<span>ϴ.
</span>
To find angle ϴ, simply take the inverse of Sinϴ= h/18... and you'll get ϴ = sin-1 (sin inverse) h/18

Hope this helps

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the answer is 50 dollars

Step-by-step explanation:

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7 - 4 ( d - 3 ) = 23

astra-53 [7]

7 - 4 ( d - 3) = 23

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The circumference of a circle can be found using the formula C = 2r. Which is an equivalent equation solved for r? r = C r = C(2

vaieri [72.5K]

Answer:

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Step-by-step explanation:

we know that

The circumference of a circle is equal to

$C=2\pi r$

where

r is the radius of the circle

Solve for r

That means -----> isolate the variable r

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Simplify right side

$\frac{C}{2\pi}=r$

Rewrite

$r=\frac{C}{2\pi}$

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r equals StartFraction C Over 2 pi EndFraction

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Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)

Yakvenalex [24]

Answer:

$\sin d = \frac{4}{7}$ ; $\sin e = \frac{\sqrt{33} }{7}$

$\cos d = \frac{\sqrt{33} }{7}$ ; $\cos e = \frac{4}{7}$

$\tan d = \frac{4}{\sqrt{33} }$ ; $\tan e = \frac{\sqrt{33} }{4}$

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

$\sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\cos \alpha = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\tan \alpha = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }$

__________________________________________________

According to the question ,

1) When α (alpha) = d

$\sin d = \frac{4}{7}$

$\cos d = \frac{\sqrt{33} }{7}$

$\tan d = \frac{4}{\sqrt{33} }$

2) When α (alpha) = e

$\sin e = \frac{\sqrt{33} }{7}$

$\cos e = \frac{4}{7}$

$\tan e = \frac{\sqrt{33} }{4}$

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