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
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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! ^^