Trigonometry

(i) Let M be the mid-point of BC.
VM/10=sin60°
VM=(10*sin60°)cm

cos∠VMA=[2(10*sin60°)²-10²]/2(10*sin60°)²
cos∠VMA=1/3
∠VMA≒70.5288°
∠VMA=70.5° (corr. to 3 sig.fig)

2013-09-08 14:07:30 補充：
(ii) Area of ABC=10*10*sin60°/2
Height of VABC=(sin70.5288°)(10sin60°)

Volume of VABC=(1/3)(10*10*sin60°/2)(sin70.5288°)(10sin60°)
=118cm³ (corr. to 3 sig.fig)

2013-09-21 20:54:02 補充：
Let M be the mid-point of BC.(i) VM/10=sin60°
VM=(10*sin60°)cmcos∠VMA=[2(10*sin60°)²-10²]/2(10*sin60°)²
cos∠VMA=1/3
∠VMA≒70.5288°
∠VMA=70.5° (corr. to 3 sig.fig) ∴the required angle is 70.5° .
(ii) Area of ΔABC=10*10*sin60°/2
Height of VABC=(sin70.5288°)(10sin60°)Volume of VABC=(1/3)(10*10*sin60°/2)(sin70.5288°)(10sin60°)
=118cm³ (corr. to 3 sig.fig)
(i) In ΔVPM,
∠PVM=(180°-70.5288°)/2
∠PVM=54.7356°PM²=x²+(10*sin60°)²-2(x)(10*sin60°)cos54.7356°
PM²=x²+75-10x
PM=√(x²+75-10x)


(ii) the required distance is from the mid-point of VA to M.

the required distance
=√(5²+75-10*5)
=7.07cm (corr. to 3 sig.fig)


The required angle
=sin^(-1)[5/(10*sin60°)]
=35.3° (corr. to 3 sig.fig)

Well Done!!!

=^o^=