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)

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)

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)

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.

Need steps,plz!

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Lv 7
5 years ago

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]

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)

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°