DAT Prep Help I S NEXT Your answer is partially correct. Try again. The drawing

Question

DAT Prep Help I S NEXT Your answer is partially correct. Try again. The drawing shows a person looking at a building on top of which an antenna is mounted. The horizontal distance between the person's eyes and the building is 95.0 m. In part a the person is looking at the base of the antenna, and his line of sight makes an angle of 35.20 with the horizontal. In part b the person is looking at the top of the antenna, and his line of sight makes an angle of 38.9 with the horizontal. How tall is the antenna? 38.9 35.2 95.0 m 95.0 m licy 1 e2000-2016

Explanation / Answer

1)    height of building above person's eye =H
height of antenna = h
a) tan (35.2) = H/95
b) tan (38.9) = (H+h)/95
h = 95 tan (38.9) - H
h = 95 [tan (38.9) - tan (35.2)]
h = 95 [0.807 - 0.705]
h = 9.69 meter

2) Draw ABC. Using the E-W straight line through A, we know 54.5° + BAC + 75.5° = 180°, so BAC = A = 180 - (54.5 + 75.5) = 50.0°.
Using the Law of Sines, sin C ÷ AB = sin A ÷ BC ==> sin C = (AB/BC) sin A = (61.9 / 96.5) sin 50.0°

==> C = 29.43°.
The third angle B = 180° - A - C = 100.57°.
Using the Law of Sines again, AC ÷ sin B = BC ÷ sin A

==> AC = BC (sin B / sin A) = 96.5(sin 100.57 / sin 50.0) = 123.8 m

3) Well for the magnitude you use good old pythagoras:   a^2 = b^2 + c^2

or a= sqrt (162^2 + 131^2) = 208.34 m

For the direction you can use good old trigonometric functions,we have here the opposite and adjacent components so you use the TAN function:
tan@= opp/adj = x/y
tan@= 162/131 = 1.237
@ = 51.04 deg
But as these are both on the negative axis it puts us in the third quadrant so you must add 180 deg.
So @ = 231.04 deg

4) Distance traveled in 0.5 hours = 95.4 * 0.5 = 47.7 km

[a] 47.7 (cos 21.6° - cos 40°) = 7.81 km

[b] 47.7 (sin 40° - sin 21.6°) = 13.10 km

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.