Triangle problem

In triangle ABC,AB = 7 cm, BC= 5 cm and AَBC = 136ْ .

http://img27.imageshack.us/img27/2061/34515568.jpg


(i) Calculate the lenght of AC.
(ii) Calculate the area of the triangle ABC.


plz
tell me hw to solve these question

Hello! I shall do part (a) first...

Have you learnt Cosine Rule?
Basically Cosine Rule states that (in the context of this question):

$\displaystyle AC^2 = AB^2 + BC^2 - 2 (AB)(BC) cos \angle ABC$

Therefore, by applying Cosine Rule, we can find AC:

$\displaystyle AC^2 = (7)^2 + 5^2 - 2 (7)(5) cos 136$

$\displaystyle AC = \sqrt{124.35} $ [rounded off to 5 significant figures] = 11.2 cm [rounded off to 3 significant figures]

Okay now for part (b)...

The formula for area of triangle is (in the context of the question):

Area of triangle ABC$\displaystyle = \frac{1}{2} (AB)(BC) sin \angle ABC$

In general, the formula is Area of triangle $\displaystyle ABC = \frac{1}{2}ab sin C$, where a and b are adjacent sides of a triangle and C is the angle opposite side c (i.e. the angle between sides a and b).

By applying this formula, we can answer the question:

Area of triangle ABC =$\displaystyle \frac{1}{2} (7)(5) sin 136$ = $\displaystyle 12.2 cm^2 $ (rounded off to 3 significant figures)

Hope this helps! ^^