Five Trigonometric Problems Required in Senior Two Mathematics. Thank you.

4.B 2-bc-2C 2 = 0 divided by BC: b/c- 1-2c/b=0.

Let b/c=x, then x- 1-2/x=0, and both b and c are side lengths, not 0, so multiply by x at the same time to get:

X 2-X-2 = 0, so x=2 (minus), which means b=2c.

According to the cosine theorem, a 2 = b 2+c 2-2bc * COSA is substituted into b=2c, a = √ 6, and COSA = 7/8:

3c^2= 12,c=2,b=4.

CosA=7/8, so sinA=√ 15/8,

So the area s =1/2bcsina = √15/2, so choose a.

5. The area S= 1/2bcsinA= 10√3, so bc=40.

Perimeter =a+b+c=20, so b 2+c 2+2bc = (20-a) 2.

That is, b 2+c 2 = (20-a) 2-80.

It can also be obtained from cosine theorem: a 2 = b 2+c 2-2bc * COSA.

Substitution gives a 2 = (20-a) 2-80-2 * 40 * 1/2.

That is, a=7. So I chose C.

6. The area S =1/2accsinb = 3/2, so ac=6.

2b=a+c, squared: 4b 2 = a 2+c 2+2ac, that is, 4b 2 = a 2+c 2+ 12.

Cosine theorem shows that B 2 = A 2+C 2-2AC * COSB means B 2 = A 2+C 2-6 √ 3.

Subtract the two formulas to get 3b 2 =12+6 √ 3.

B=√3+ 1, so choose B.