A boy 1.5 m tall is standing 12 m away from a church building which has a tower on top of its roof. The top of the cross on the tower is 14.6 m away from the boy’s head (eyes). If the boy has to raise his eyes through an angle of 31° in order to see the top of the roof, calculate, Correct to the nearest degree, the angle through which the boy must raise his eyes to see the top of the cross on the tower. Correct to one decimal place, the height of the top of the cross from ground.
Correct to one decimal place the height of the church building.

1.

2. 9.8 m

3. 8.7 m.

Step-by-step explanation:

1. Let the angle be represented by θ, so that:

Cos θ =

          =

Cos θ = 0.82192

θ = 0.8219

   =

θ =

The angle through which the boy must raise his eyes to see the top of the cross on the tower is .

2. Let the height from the boy's head to the top of the cross be represented by y, so that:

Sin =

y = Sin x 14.6

  = 0.5697 x 14.6

y = 8.32 m

The height of the top of the cross from the ground = 1.5 + 8.32

                                                                 = 9.82 m

The height of the top of the cross from the ground is 9.8 m.

3. Let the height from the boy's head to the top of the building be represented by x, so that:

Tan 31 =

            =

x = Tan 31 x 12

  = 0.6009 x 12

x = 7.21

The height of the church building = 1.5 + 7.21

                                           = 8.71

The height of the church building is 8.7 m.

The sketch to the question is attached to this answer.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: