Question

un escalier mobile mesure 14,5 mètres. si langle dinclinaison mesure 27°, détermine la hauteur de cet escalier mobile au dixième de mètre près.
la distance entre deux édifices est de 25 mètres. à partir du toit du plus petit édifice, langle délévation jusquau toit de lautre édifice est de 31° et langle de dépression jusquà la base de lautre édifice est de 48°. détermine la hauteur du plus grand édifice. arrondis au dixième près.

Explanation:

For the first problem (Escalier mobile):

Step1: Identify the triangle type

We have a right triangle \( ABC \) (since \( \angle C = 90^\circ \)), where the length of the escalier (hypotenuse \( AB \)) is \( 14.5 \) m and the angle of inclination \( \angle A = 27^\circ \). We need to find the height \( BC \) (opposite to \( \angle A \)).

We use the sine function: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)

So, \( \sin(27^\circ) = \frac{BC}{14.5} \)

Step2: Solve for \( BC \)

Multiply both sides by \( 14.5 \):

\( BC = 14.5 \times \sin(27^\circ) \)

Calculate \( \sin(27^\circ) \approx 0.4540 \)

Then \( BC \approx 14.5 \times 0.4540 \approx 6.583 \approx 6.6 \) (rounded to the nearest tenth)

Step1: Find the height of the shorter building (\( h_1 \))

Using the angle of depression \( 48^\circ \) and the distance \( d = 25 \) m. The angle of depression is equal to the angle of elevation from the base of the taller building to the top of the shorter building. So:

\( \tan(48^\circ)=\frac{h_1}{25} \)

\( h_1 = 25\times\tan(48^\circ) \)

\( \tan(48^\circ)\approx1.1106 \), so \( h_1\approx25\times1.1106 = 27.765 \) m

Step2: Find the height from the top of the shorter to the taller (\( h_2 \))

Using the angle of elevation \( 31^\circ \) and \( d = 25 \) m:

\( \tan(31^\circ)=\frac{h_2}{25} \)

\( h_2 = 25\times\tan(31^\circ) \)

\( \tan(31^\circ)\approx0.6009 \), so \( h_2\approx25\times0.6009 = 15.0225 \) m

Step3: Find the height of the taller building (\( H \))

\( H=h_1 + h_2\approx27.765+15.0225 = 42.7875\approx42.8 \) m

Answer:

The height of the escalator mobile is approximately \( 6.6 \) meters.

For the second problem (Deux édifices):

Let's denote the height of the shorter building as \( h_1 \) and the taller one as \( h_2 \), and the distance between them as \( d = 25 \) m.

For the angle of elevation \( 31^\circ \) (to the top of the shorter building):

\( \tan(31^\circ) = \frac{h_1}{25} \)

\( h_1 = 25 \times \tan(31^\circ) \approx 25 \times 0.6009 \approx 15.0225 \) m

For the angle of depression \( 48^\circ \) (to the base of the taller building, which is equivalent to an angle of elevation of \( 48^\circ \) from the base of the taller building to the top of the shorter building? Wait, actually, the angle of depression from the top of the shorter building to the base of the taller building is \( 48^\circ \), so the height from the base of the taller building to the top of the shorter building is \( h_1 \), and the horizontal distance is \( d = 25 \) m. Wait, maybe I misinterpret. Let's re - explain:

Wait, the distance between the two buildings is \( d = 25 \) m. Let the height of the shorter building be \( h \) and the taller one be \( H \).

From the top of the shorter building, the angle of depression to the base of the taller building is \( 48^\circ \), so the vertical distance from the top of the shorter building to the base of the taller building is \( h \) (since the angle of depression is equal to the angle of elevation from the base of the taller building to the top of the shorter building). So \( \tan(48^\circ)=\frac{h}{25} \), so \( h = 25\times\tan(48^\circ)\approx25\times1.1106\approx27.765 \) m? No, that can't be. Wait, no, the angle of elevation to the top of the shorter building is \( 31^\circ \), so the height of the shorter building \( h_1 = 25\times\tan(31^\circ)\approx15.02 \) m. The angle of depression from the top of the shorter building to the base of the taller building is \( 48^\circ \), so the height from the base of the taller building to the top of the shorter building is \( h_1 \), and the height of the taller building \( H=h_1 + h_3 \), where \( h_3 \) is the height from the top of the shorter building to the top of the taller building? Wait, no, maybe the angle of elevation to the top of the taller building is \( 31^\circ \)? Wait, the problem says: "À partir du toit du plus petit édifice, l'angle d'élévation jusqu'au toit de l'autre édifice est de \( 31^\circ \) et l'angle de dépression jusqu'à la base de l'autre édifice est de \( 48^\circ \)".

So:

1. Angle of depression \( 48^\circ \) to the base of the taller building: this means that the height of the shorter building \( h_1 \) is given by \( \tan(48^\circ)=\frac{h_1}{25} \) (because the horizontal distance is 25 m, and the angle of depression is equal to the angle of elevation from the base of the taller building to the top of the shorter building).

\( h_1 = 25\times\tan(48^\circ)\approx25\times1.1106 = 27.765 \) m

1. Angle of elevation \( 31^\circ \) from the top of the shorter building to the top of the taller building: let the height from the top of the shorter building to the top of the taller building be \( h_2 \). Then \( \tan(31^\circ)=\frac{h_2}{25} \)

\( h_2 = 25\times\tan(31^\circ)\approx25\times0.6009 = 15.0225 \) m

Then the height of the taller building \( H=h_1 + h_2\approx27.765 + 15.0225 = 42.7875\approx42.8 \) m (rounded to the nearest tenth)