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A 10-foot ladder placed on level ground leans against the side of a house. The ladder reaches a point that is 9.2 feet up on the side of the house. ( Answer options A and B and show your work). Will Mark Brainliest if answered correctly. ( Do Not Repost Someone Else's Answers That's Already On Brainly And Any Other Websites Or You'll Be Reported). If you can't help please do not leave a response thank you. ​

A 10foot Ladder Placed On Level Ground Leans Against The Side Of A House The Ladder Reaches A Point That Is 92 Feet Up On The Side Of The House Answer Options A class=

Sagot :

So

Let's see

  • sinA=Perpendicular/Hypotenuse
  • sinA=9.2/10
  • sinA=0.92
  • A=sin^{-1}(0.92)
  • A=66.9°

#B

So

We need base

  • cosA=Base/Hypotenuse
  • cos75=B/10
  • B=10cos75
  • B=2.58ft
View image Аноним

Answer:

a) 67°

b) 2.6 ft

Step-by-step explanation:

Part (a)

This problem can be modeled as a right triangle (see attachment 1), where the wall and the ground are the legs, and the ladder is the hypotenuse. Therefore, we can use the sine trig ratio to find the angle.

Trig Ratio

[tex]\sf \sin(\theta)=\dfrac{O}{H}[/tex]

where:

  • [tex]\theta[/tex] is the angle
  • O is the side opposite the angle
  • H is the hypotenuse (the side opposite the right angle)

Given:

  • [tex]\theta[/tex] = x
  • O = wall = 9.2
  • H = ladder = 10

Substituting the given values into the ratio and solving for x:

[tex]\implies \sf \sin(x)=\dfrac{9.2}{10}[/tex]

[tex]\implies \sf x=\sin^{-1}\left(\dfrac{9.2}{10}\right)[/tex]

[tex]\implies \sf x=67^{\circ}\:(nearest\:degree)[/tex]

Part (b)

(see attachment 2)

Let y be the distance the foot of the ladder and the foot of the building

We can find y by using the cos trig ratio:

[tex]\sf \cos(\theta)=\dfrac{A}{H}[/tex]

where:

  • [tex]\theta[/tex] is the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

Given:

  • [tex]\theta[/tex] = 75°
  • A = grounds = y
  • H = ladder = 10

Substituting the given values into the ratio and solving for x:

[tex]\implies \sf \cos(75^{\circ})=\dfrac{y}{10}[/tex]

[tex]\implies \sf y=10\cos(75^{\circ})[/tex]

[tex]\implies \sf y=2.6\:ft\:(nearest\:tenth)[/tex]

View image semsee45
View image semsee45
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