ni asked in 科學及數學數學 · 5 years ago

4 F3 Maths Problems, Quick !

1) In the figure, a rectangle box ABCD is 2m high and 1m wide. The box stands on a board EB which slopes upward with gradient 1:2. If EF is the level ground, ∠BFE=90° and EF=4m, find the height of C above the ground.(Give the answer correct to 3 significant figures)

Figure:http://postimg.org/image/ukmjp4ma9/

2)In the figure, AB is an antenna standing upright on top of a radar station BC. D is a point on the horizontal ground. The angles of elevation of A and B from D are 25° and 15° respectively. E is a point between D and C such that DE=12m. If the angle of depression of E from A is 75°, find

(a) the height BC of the radar station,

(b) the height AB of the antenna.

(Give the answers correct to the nearest 0.1m)

Figure:http://postimg.org/image/5yqvt3233/

3)The Bermuda Triangle is a mysterious region in the Altanic Ocean. Many aeroplanes and ships have mysteriously gone missing there for the region region. Suppose AB=1700km, ∠BAC=40°, ∠ACB=24° and the bearing of C from B is S54°E.

(a) What is the bearing of point A from point B?

(b) Let h km be the distance between Bermuda Islnad and AC. Find the value of h.

(c) Find the area of the Bermuda Triangle .

(Given the answers correct to 3 significant figures if necessary)

Figure:http://postimg.org/image/tpjiivkfb/

4)In the figure, A and B are two piers 5km apart such that B is due east of A. boats, P and Q leave A and B respectively at the same time into the open sea at the same speed. P moves on a course 145° while that of Q is 250°.

(a) If P and Q maintain the same speed and direction throughtout the course, will they clash eventually? give reasons.

(b) Write down a set of possible bearings for P and Q which will cause them for clash.

figure:http://postimg.org/image/9fhtzh2e3/

Need steps,plz!

1 Answer

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  • 土扁
    Lv 7
    5 years ago
    Favorite Answer

    1.

    tan∠BAC = 2 and tan∠EBF= 2

    ∠BAC = ∠EBF

    CA // BF (alt. ∠s equal)

    CA is a vertically line.

    CA² = AB² +BC² (Pythagorean theorem)

    CA² = 2² + 1² m²

    CA = √5 m

    Let h m be the height of A above the ground.

    4² = h² + (2h)² (Pythagorean theorem)

    h = 4(√5)/5

    The height of C above the group

    = (√5) + 4(√5)/5 m

    = 4.02 m (to 3 sig. fig.)

    ====

    2.

    (a)

    Let AC = y m, and EC = x m

    In ΔAEC :

    tan75° = y/x

    y = x tan75° ...... [1]

    In ΔADC :

    tan25° = y/(12 + x)

    y = (12 + x) tan25° ...... [2]

    [1] = [2] :

    x tan75° = (12 + x) tan25°

    x (tan75° - tan25°) = 12 tan25°

    x = 12 tan25° / (tan75° - tan25°)

    In ΔDBC :

    tan15° = BC / {12 + [12 tan25° / (tan75° - tan25°) m]}

    BC = tan15° * {12 + [12 tan25° / (tan75° - tan25°)]} m

    BC = 3.7 m (to the nearest 0.1m)

    (b)

    [1] :

    y = [12 tan25° / (tan75° - tan25°)] * tan75°

    y = 6.4

    AB

    = 6.4 - 3.7 m

    = 2.7 m (to the nearest 0.1m)

    ====

    3.

    (a)

    bearing

    = S (180° - 54° - 40° - 24°) W

    = S 62° W

    (b)

    In ΔABD :

    sin40° = h / 1700

    h = 1700 sin40°

    h = 1090 (to 3 sig. fig.)

    (c)

    In ΔABD :

    cos40° = AD / (1700 km)

    AD = 1700 cos40° km

    In ΔBDC :

    tan24° = (h km) / DC

    DC = h / tan24° km

    DC = 1700 sin40° / tan24° km

    Area of ΔBAC

    = (1/2) × [1700 cos40° + (1700 sin40° / tan24°)] × (1700 sin40°) km²

    = 2050000 km²

    ====

    4.

    (a)

    Let both P and Q pass through point C (may not at the same time).

    Draw CD ⊥ AB and cut AB at D.

    In ΔACD :

    sin(145° - 90°) = CD/AC

    AC = CD / sin55°

    In ΔBCD :

    sin(270° - 250°) = CD/BC

    BC = CD / sin20°

    Obviously, CD / sin55° ≠ CD / sin20°, then AC ≠ BC

    As P and Q have the same speed, the time needed for they arrived at C isdifferent.

    Thus, theywill NOT clash eventually.

    (b)

    As, P and Q have the same speed, they will clash when AC = BC.

    When AC = BC, the isosceles triangle would have equal base angles, ∠CAB = ∠CBA

    When ∠CAB = ∠CBA= 45° :

    Bearing for P = 90° + 45° = 135°

    Bearing for Q = 270° - 45° = 225°

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