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Luda [366]
3 years ago
8

El angulo de elevacion del sol es de 29°12'35''. Calcular la longitud de la sombra proyectada por un hombre de 1,75m de estatura

.
Mathematics
1 answer:
Bingel [31]3 years ago
3 0

Respuesta:

3,59 metros

Explicación paso a paso:

Conversión del ángulo de elevación a grados;

29 ° 12 '35 "

= 29 ° + 12 '/ 60 + 35 "/ 3600

= 29,20972 °

Resolviendo el triángulo de solución adjunto:

La longitud de la sombra se puede obtener utilizando la relación trigonométrica;

Sin θ = opuesto / hipoteno

Sin 29.20972 = 1.75 / sombra

Sombra = 1.75 / Sin 29.20972

Longitud de la sombra = 3,59 m

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How to solve part ii and iii
iragen [17]

(i) Given that

\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}

when x=1 this reduces to

\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

2 \tan^{-1}(y) = \dfrac\pi3

\tan^{-1}(y) = \dfrac\pi6

\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)

\implies \boxed{y = \dfrac1{\sqrt3}}

(ii) Differentiate \tan^{-1}(xy) implicitly with respect to x. By the chain and product rules,

\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}

(iii) Differentiating both sides of the given equation leads to

\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0

where we use the result from (ii) for the derivative of \tan^{-1}(xy).

Solve for \frac{dy}{dx} :

\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0

\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)

\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}

\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}

\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}

From part (i), we have x=1 and y=\frac1{\sqrt3}, and substituting these leads to

\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}

\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}

\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}

as required.

3 0
2 years ago
How do I solve and graph this
White raven [17]
I hope this helps you out

5 0
4 years ago
paige works at a deployment store and has a commission rate of 3 percent. her sales for the quarter were $4596. if she earns $50
Afina-wow [57]

Solution-

∵ she earns $50 per day and she worked 30 days in the quarter.

∴ Her total earning for this period = $(50×30) = $1500.

Also she earns 3% commission for her sales.

∴ Her total earning by commission =3% of $4596= $ (3×4596)/100

                                                           = $ 137.88

∴Total amount earned by Paige= $137.88+$1500

                                                     =$ 1637.88

5 0
3 years ago
Perrie invests £25000 for 3 years in a saving account. She gets 2.7% per annum compound interest. Calculate the total amount of
aleksandr82 [10.1K]

Answer:

I (interest) = £ 2,080.17

Step-by-step explanation:

Perrie invests £25000 for 3 years in a saving account. She gets 2.7% per annum compound interest. Calculate the total amount of interest perrie will get after 3 years.

To solve the above question, we would use the Periodic compound interest formula

A = P(1 + r) ^t

Where P = Principal = £25000

r = Interest rate = 2.7 % = 0.027

t = time in years = 3 years

Hence:

A = £25,000( 1 + 0.027)³

A = £27,080.17

A = P + I where

P (principal) = £ 25,000.00

I (interest) = £27,080.17 - £ 25,000.00

I (interest) = £ 2,080.17

The total amount of interest perrie will get after 3 years is £ 2,080.17

3 0
3 years ago
Find a solution to 3cos(θ)=−0.862 in radians
eimsori [14]
3cos(\theta) = -0.862
\frac{3cos(\theta)}{3} = \frac{-0.862}{3}
cos(\theta) = -\frac{431}{1500}
cos^{-1}[cos(\theta)] = cos^{-1}(-\frac{431}{1500})
\theta = 106.6983737\°
\theta = 1.86223792^{r}
7 0
3 years ago
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