In a triangle ABC, the side AB is 10 cm and the side AC is 8 cm. The angle BAC is 45 degrees.?

Question

i dont know how to find

(i) Find the length BC.
(ii) Find the angle ABC .

Answer 1

The Law of Cosines states:
C=(a^2+b^2+abCos C)^-2
C=(10^2+8^2+10*8*Cos45)^-2
C=(100+64+80(0.707))^-2
C=(220.57)^-2=14.85 approximately, which is the length of BC

The Law of Sines states:
a/Sin A=b/SinB=c/Sin C
a/Sin A = 45(degrees)/14.85 = X (degrees)/10,
X (in degrees) = 45 (degrees)*10/14.85 = 66.8 (degrees)

Answer 2

If its a right triangle- angle ABC must be 45 degrees also because total angle of any triangle is 180
therefore the two short sides are equal in length so length bc is also 8 cm

Answer 3

Use the formula a^2=b^2 + c^2 - 2bc*cosA (i hope you understand this formula. if not, check the given source), you get BC = 7.133 cm .

Use the same formula, that is, cosB=(c^2 + a^2 - b^2)/ 2ca
This gives you the angle ABC to be 52.48 degrees.