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2 years ago

Use the diagram below to answer the question.

Mathematics

1 answer:

Lunna [17]2 years ago

7 0

Answer:

The length of shadow is 69 foot.

Step-by-step explanation:

We are given a shadow of lighthouse.

Length of top of lighthouse to end of shadow is 90 ft.

Angle make at top of lighthouse is 50°

Distance from foot of lighthouse to end of shadow is x ft.

we need to find the length of shadow i,e x ft

Using trigonometry ration:

$\sin\theta=\dfrac{P}{H}$

$\sin50^\circ=\dfrac{x}{90}$

$x=90\times 0.766$

$x=68.94\approx 69\text{ ft}$

Hence, The length of shadow is 69 foot.

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In ΔKLM, the measure of ∠M=90°, the measure of ∠K=78°, and KL = 35 feet. Find the length of MK to the nearest tenth of a foot.

Margaret [11]

Answer:

7.3 feet

Step-by-step explanation:

osK=

hypotenuse

adjacent

\cos 78=\frac{x}{35}

cos78=

35\cos 78=x

35cos78=x

Cross multiply.

x=7.2769\approx \mathbf{7.3}\text{ feet}

x=7.2769≈7.3 feet

Type into calculator and round to the nearest tenth of a foot.

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What is the distance between (9,5) and (-2,2)

nikitadnepr [17]

Answer:

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Step-by-step explanation:

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A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .