C2 Qs on sine and cosine rule. Pls help me out!!!

Hey, could somebody pls help me with the following questions? (pg. 32 of C2 book) Thanks, I'd really appreciate it! :redface:

9) In triangle ABC, AB=sqrt 2 cm, BC= sqrt 3 cm, and angle BAC=60. Show that angle ACB=45 and find AC.

10) In triangle ABC, AB= (2-x)cm, BC= (x+1)cm and angle ABC=120.
a) show that AC^2=x^2-x+7
b) find the value of x for which AC has a minimum value.

11) Triangle ABC is such that BC=5sqrt2 cm, angle ABC=30 and angle BAC=theta (cant paste symbol), where sin theta=sqrt5 / 5.
Work out the length of AC, giving your answer in the form AsqrtB, where A and B are integers.

12) The perimeter of triangle ABC=15 cm. Given that AB=7 cm, and angle BAC=60, find the lengths of AC and BC.

Reply 1

evariste_3086

1

*girlie*

Hey, could somebody pls help me with the following questions? (pg. 32 of C2 book) Thanks, I'd really appreciate it! :redface:

9) In triangle ABC, AB=sqrt 2 cm, BC= sqrt 3 cm, and angle BAC=60. Show that angle ACB=45 and find AC.

10) In triangle ABC, AB= (2-x)cm, BC= (x+1)cm and angle ABC=120.
a) show that AC^2=x^2-x+7
b) find the value of x for which AC has a minimum value.

11) Triangle ABC is such that BC=5sqrt2 cm, angle ABC=30 and angle BAC=theta (cant paste symbol), where sin theta=sqrt5 / 5.
Work out the length of AC, giving your answer in the form AsqrtB, where A and B are integers.

12) The perimeter of triangle ABC=15 cm. Given that AB=7 cm, and angle BAC=60, find the lengths of AC and BC.

9) sin60/sqrt(3)=sin(ACB)/sqt(2)

1/2.sqrt(2)=sin (ACB)
so ACB=45 so angle ABC is (180-45-60) then use sine rule again to find other side using sinABC/AC=sin60/sqrt(3)

10 use cos rule
ac^2=ab^2+cb^2-2(ab)(cb)cos(abc)
=(2-x)^2+(x+1)^2-2(2-x)(1+x)(-1/2)
=4-4x+x^2+x^2+2x+1+2+x-x^2
=x^2-x+7
to minimise diff wrt to x and set to 0
so 2x-1=0 then x=1/2
diif again to get 2 >0 so this value of x is a min.
so AC^2 min when x=1/2

11)
AC/sin 30=BC/sin theta
AC=1/2.5sqrt(2)/sqt(5)/5=25sqrt(2)/2sqrt(5)=5sqrt(10)/2...not in the form as stated either ive made a slip or question not correct ill let you check my working.

12) ab+bc+ac=15
ab=7
so bc+ac=8...(1)
by cos rule
bc^2=7^2+ac^2-2.7.ac.cos(120)
bc^2=49+ac^2+7ac
using 1 bc=(8-ac)
so
64-16ac+ac^2=49+ac^2+7ac
64-49=7ac+16ac
which gives ac and so the other side can be calculated. once again not a nice answer, apologies for any numerical errors, its hard doing this at computer !!

Reply 2

*girlie*

OP

5

thanks!

the answer to Q11 should be 4sqrt10...hmmmmm

C2 Qs on sine and cosine rule. Pls help me out!!!

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