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Shkiper50 [21]
2 years ago
15

A 6-foot ladder touches the side of a building at a point 5 feet above the ground. At what height would a 18-foot ladder touch t

he building if it makes the same angle with the ground as the shorter ladder?
Mathematics
1 answer:
vovangra [49]2 years ago
6 0

Answer:

15 feet above the ground

Step-by-step explanation:

We know sinθ = opposite/hypotenuse where h = height of the ladder above the ground = opposite = 5ft and L = length of ladder = hypotenuse side = 6 feet. sinθ = h/L

For the new ladder to make the same angle, sinθ must be the same, so let h' = height of new ladder above the ground and L' = length of new ladder = 18 ft

sinθ = h'/L'

So. h'/L' = h/L

h' = hL'/L

substituting the values of the variables, we have

h' = 5 ft × 18 ft/6 ft

= 5 ft × 3

= 15 ft

So, the 18-foot ladder must be placed 15 feet above the ground to make the same angle as the smaller ladder.

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ahmed is sitting on a dock watching a float plane in a harbour. At a certain time, the plane is 350 m above the water and 470 m
Pavel [41]

Answer:

The angle of elevation of the plane measured from Ahmed is approximately 36.67°

Step-by-step explanation:

The given parameters of the float plane are;

The height of the float plane above the water, y = 350 m

The horizontal distance of the float plane from Ahmed, x = 470 m

Given that Ahmed is sitting on the dock, by the water, by trigonometric ratios, we have;

The height of the float plane, the distance of the plane from Ahmed and the line of sight forming the angle of elevation of the plane measured from Ahmed, form a right triangle

tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent \ Leg\ length}

Therefore

tan(\theta) = \dfrac{y}{x}

Where;

θ = The angle of elevation of the plane measured from Ahmed

y = The leg of the right triangle opposite the reference angle

x = The leg  of the right adjacent to reference angle

Therefore;

θ = arctan(y/x) which gives;

θ = arctan(350/470) ≈ 36.67°

The angle of elevation of the plane measured from Ahmed, θ ≈ 36.67°.

8 0
3 years ago
Given: Right triangles CES and RST. In two or more complete sentences, explain what additional information is needed to prove tr
masya89 [10]
You know one angle is 90 degrees because they are right triangles. In order to prove CES congruent to RST, you would need two legs, three angles, or a leg and an included angle.
7 0
2 years ago
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The shapes below are mathematically similar​
s2008m [1.1K]

Answer:

5 cm is the answer brochacho

7 0
2 years ago
25 points please help thanks
weeeeeb [17]

Answer:

Step-by-step explanation:

With a slope of 1 and a y intercept of -3

y = 1x - 3

y = 1(-2) - 3

y = -5

(-2, -5)

6 0
2 years ago
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
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