F.4 Maths trigonometry

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2014-05-03 13:57:45 補充：
咁就唔好COPY啦~

話說扣點數呢家野真係...

2014-05-03 15:03:16 補充：
a)
Since M is the mid-pt of QR
MQ=MR=3 cm
Also, we know that ∠QMR=50°.

So,
by the cosine formula,
QR^2 = MQ^2 + MR^2 - 2(MQ)(MR) cos∠QMR
QR^2 = 3^2 + 3^2 - 2(3)(3) cos 50°
QR^2 = 18 - 18 cos 50°
QR = 2.53570957 cm
QR = 2.54 cm ( cor to 3 sig fig )

b)
For the triangle MQR,
we know that MQ=MR=3 cm , ∠QMR=50°

Therefore,
The area of triangle MQR
=(1/2)(MQ)(MR) sin∠QMR
=(1/2)(3)(3) sin 50°
=(9/2) sin 50°
=3.45 cm^2 ( cor to 3 sig fig )

For the triangle PQR,
we know that PQ=PR=5 cm , QR=2.53570957 cm

s = (5+5+2.53570957)/2 = 6.267854785 cm

By the Heron's formula,
The area of triangle PQR
=√[ (6.267854785)(6.267854785-5)(6.267854785-5)(6.267854785-2.53570957) ]
=6.132086096 cm
=6.13 cm^2 ( cor to 3 sig fig )

c)
Now, we know the areas of triangle PQR and triangle MQR.
Let h be the shortest distance from M to the horizontal surface.

PM^2 =PR^2 - MR^2
PM = √(5^2 - 3^2) = √16 = 4 cm

For the volume of tetrahedron MPQR,
volume =(1/3)(Area of triangle PQR)(h) ... 1
Also, the volume = (1/3)(Area of triangle MQR)(PM) ... 2
Combine 1 and 2,
(1/3)(Area of triangle PQR)(h) = (1/3)(Area of triangle MQR)(PM)
(Area of triangle PQR)(h) = (Area of triangle MQR)(PM)
6.132086096 h = [(9/2) sin 50°] (4)
h = 2.248631177 cm
h = 2.25 cm ( cor to 3 sig fig )
Therefore, the shortest distance from M to the horizontal surface is 2.25 cm.

d) Let N be the mid-pt of QR.
In triangle MNR and triangle MNQ,
MR=MQ ... given
MN=MN ... common
QN=RN
Therefore, triangle MNR ~= triangle MNQ. ( SSS )
So ∠NMQ=∠NMR=50°/2=25° ... ( corr ∠s, ~= tiangles )
MN=MR cos 25°
MN = 2.718923361 cm

Let θ be the angle between the plane MQR and the horizontal surface.
sin θ = h / MN
sin θ = 2.248631177 / 2.718923361
θ=55.8° ( cor to 3 sig fig )
Therefore, the angle between the plane MQR and the horizontal surface is 55.8°.

e)
因為字數問題，請到意見區看看。

2014-05-03 15:03:42 補充：
e)
When ∠QMR increases from 50° to 90°, the area of triangle MRQ will increase.
Then the volume will also increase because volume = (1/3)(Area of triangle MRQ)(PM) where PM is fixed.
When ∠QMR increases from 90° to 120°, the area of triangle MRQ will decrease.
Then the volume will also decrease.

(a) QR^2=3^2+3^2-(3)(2)cos50degree
QR=2.54cm (to 3 s.f.)
(b) areas of triangle MQR=1/2*3*3*sin50
=3.45cm^2 (to 3 s.f.)
s=1/2*(5+5+6)=8cm
so, Area of triangle PQR=√8*(8-5)*(8-5)*(8-6)
=12cm^2
(c) consider triangle MQR (from (b)) and QR (from (a))
1/2*3*3*sin50=1/2*(√3^2+3^2-(3)(2)cos50degree)*required shortest distance
required shortest distance=2.716535...cm
(d) ********************************************************************************************
(e) ********************************************************************************************

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