Thread: find lengths of triangle's 2 sides

hi

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

Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..

Originally Posted by AbYz

hi

The perimeter of triangle ABC = 15 cm. Given that AB = 7 cm and angle BAC = 60 degreez, find the lengths of AC and BC.
Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..

The rule of cosines tells us that
$(8-b)^2=b^2+49-7b\cos(60^o)$

Originally Posted by Plato

The rule of cosines tells us that
$(8-b)^2=b^2+49-\color{red}{2} \cdot 7b\cos(60^o)$

...

method 2 ... use Pythagoras on the larger right triangle

Originally Posted by AbYz

Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..

You didn't know that COS(60) = 1/2 ?

Originally Posted by DenisB

You didn't know that COS(60) = 1/2 ?

somehow i ruled out using sine/cosine rules at the beginning..