From a plane due east at 265 m above sea level, the angle of depression of two ships sailing due east measure 35° and 25°.how far apart are the ships
✏️TRIGONOMETRY
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[tex] \underline{\mathbb{PROBLEM}:} [/tex]
- From a plane due east at 265 m above sea level, the angle of depression of two ships sailing due east measure 35° and 25°. How far apart are the ships?
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[tex] \underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad\Large » \:\tt\purple{\approx 189.83 \: meters} [/tex]
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[tex] \underline{\mathbb{SOLUTION}:} [/tex] First, we need to determine the distance of the ships from the foot of the altitude of the plane.
• For Ship A, the angle of depression from the plane is 35°, complemented to 55°. This angle would be our reference in which it is adjacent to the altitude of the plane and opposite to the distance between the altitude of the plane and Ship A.
- [tex] \tan \theta = \frac{opposite}{adjacent} \\ [/tex]
- [tex] \tan 55 \degree = \frac{opp.}{265\,m} \\ [/tex]
- [tex] opp. = 265\,m(\tan 55 \degree) [/tex]
- [tex] opp. \approx 378.46\,m [/tex]
» Therefore, the distance between the altitude of the plane and Ship A is about 378.46 meters. Find the distance for ship B.
- [tex] \tan \theta = \frac{opposite}{adjacent} \\ [/tex]
- [tex] \tan 65 \degree = \frac{opp.}{265\,m} \\ [/tex]
- [tex] opp. = 265\,m(\tan 65 \degree) [/tex]
- [tex] opp. \approx 568.29\,m [/tex]
» After finding their distance from the altitude of the plane, solve for the measure of their distances which is their differences.
- ≈ [tex] Ship \, B – Ship \, A [/tex]
- ≈ [tex] 568.29\,m – 378.46\,m [/tex]
- ≈ [tex] 568.29\,m – 378.46\,m [/tex]
- ≈ [tex] 189.83\,m [/tex]
[tex] \therefore [/tex] The distance between the two ships is approximately 189.83 meters.
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