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Sagot :
First, we will convert the given angles of elevation and depression into terms that we can use in trigonometric formulas. Note the following:
- The height of the window above the ground, [tex]\( h = 15 \)[/tex] meters.
- The angle of elevation of the top of the opposite house, [tex]\( \theta_1 = 30^\circ \)[/tex].
- The angle of depression of the foot of the opposite house, [tex]\( \theta_2 = 45^\circ \)[/tex].
### Step 1: Determine the horizontal distance between the two houses
Using the angle of depression ([tex]\(45^\circ\)[/tex]) and the height of the window ([tex]\(h\)[/tex]), we can find the horizontal distance ([tex]\(d\)[/tex]) between the two houses. The formula we use is based on the tangent of the angle of depression:
[tex]\[ \tan(45^\circ) = \frac{h}{d} \][/tex]
Since [tex]\(\tan(45^\circ) = 1\)[/tex], we have:
[tex]\[ 1 = \frac{15}{d} \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ d = 15 \text{ meters} \][/tex]
### Step 2: Determine the height of the opposite house above the window
Now, using the angle of elevation ([tex]\(30^\circ\)[/tex]), we can find the additional height ([tex]\(H\)[/tex]) of the opposite house above the level of the window. The formula we use is:
[tex]\[ \tan(30^\circ) = \frac{H}{d} \][/tex]
Since [tex]\(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)[/tex], we have:
[tex]\[ \frac{1}{\sqrt{3}} = \frac{H}{15} \][/tex]
Solving for [tex]\(H\)[/tex]:
[tex]\[ H = 15 \times \frac{1}{\sqrt{3}} \][/tex]
[tex]\[ H = 15 \times \frac{\sqrt{3}}{3} \][/tex]
Given [tex]\(\sqrt{3} = 1.732\)[/tex]:
[tex]\[ H = 15 \times \frac{1.732}{3} \][/tex]
[tex]\[ H = 15 \times 0.577 = 8.66 \text{ meters} \][/tex]
### Step 3: Calculate the total height of the opposite house
Finally, we add the height of the window ([tex]\(h = 15\)[/tex] meters) to the height of the house above the window ([tex]\(H = 8.66\)[/tex] meters) to get the total height of the opposite house:
[tex]\[ \text{Total height} = h + H = 15 + 8.66 = 23.66 \text{ meters} \][/tex]
Thus, the height of the opposite house is [tex]\(23.66\)[/tex] meters.
- The height of the window above the ground, [tex]\( h = 15 \)[/tex] meters.
- The angle of elevation of the top of the opposite house, [tex]\( \theta_1 = 30^\circ \)[/tex].
- The angle of depression of the foot of the opposite house, [tex]\( \theta_2 = 45^\circ \)[/tex].
### Step 1: Determine the horizontal distance between the two houses
Using the angle of depression ([tex]\(45^\circ\)[/tex]) and the height of the window ([tex]\(h\)[/tex]), we can find the horizontal distance ([tex]\(d\)[/tex]) between the two houses. The formula we use is based on the tangent of the angle of depression:
[tex]\[ \tan(45^\circ) = \frac{h}{d} \][/tex]
Since [tex]\(\tan(45^\circ) = 1\)[/tex], we have:
[tex]\[ 1 = \frac{15}{d} \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ d = 15 \text{ meters} \][/tex]
### Step 2: Determine the height of the opposite house above the window
Now, using the angle of elevation ([tex]\(30^\circ\)[/tex]), we can find the additional height ([tex]\(H\)[/tex]) of the opposite house above the level of the window. The formula we use is:
[tex]\[ \tan(30^\circ) = \frac{H}{d} \][/tex]
Since [tex]\(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)[/tex], we have:
[tex]\[ \frac{1}{\sqrt{3}} = \frac{H}{15} \][/tex]
Solving for [tex]\(H\)[/tex]:
[tex]\[ H = 15 \times \frac{1}{\sqrt{3}} \][/tex]
[tex]\[ H = 15 \times \frac{\sqrt{3}}{3} \][/tex]
Given [tex]\(\sqrt{3} = 1.732\)[/tex]:
[tex]\[ H = 15 \times \frac{1.732}{3} \][/tex]
[tex]\[ H = 15 \times 0.577 = 8.66 \text{ meters} \][/tex]
### Step 3: Calculate the total height of the opposite house
Finally, we add the height of the window ([tex]\(h = 15\)[/tex] meters) to the height of the house above the window ([tex]\(H = 8.66\)[/tex] meters) to get the total height of the opposite house:
[tex]\[ \text{Total height} = h + H = 15 + 8.66 = 23.66 \text{ meters} \][/tex]
Thus, the height of the opposite house is [tex]\(23.66\)[/tex] meters.
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