AS mathematics sine and cosine rule question...?

Question

Can anyone help we with this as mathematics sine and cosine rule question??
The triangle ABC with AB= x cm, BC= 3x cm, AC= 7 cm and angle B= 60°.
a) show that x = √7
b) Find angle C

Answer 1

let sides a = 3x

b = 7

c = x and B = 60*

by cosine rule b^2 = a^2 + c^2 - 2*a*c*cos B

cos 60* = 1/2

hence b^2 = a^2 + c^2 - a*c

substituting the values you will get x = sq rt(7)

now since you have x again apply cosine rule

cos C = ( a^2 + b^2 - c^2)/ (2*a*b)

cos C = 5/(2*sq rt (7))

Answer 2

The sine rule is once you have opposite over Hypotenuse and actual that is in case you're able to discover the answer to the two of those 2 lengths. Cosine is adjoining over Hypotenuse and Tangent is opposite over adjoining. that is no longer elementary to describe because of the fact i do no longer comprehend how a lot you comprehend approximately this geometric mathematical equation.

Answer 3

a) use cosine rule, so x^2 +9x^2 - 2(x)(3x)cos60 = 49. So 7x^2 = 49 or x^2 = 7 hence x = rt7.

b)B = 60. So A+C= 120 and sinA/sinC = BC/AB = 3.

So sinA = 3sinC which means sin(120-C) = 3sinc.

sin120cosc - cos120sinC = 3sinC.

[(rt3)/2]cotC - (1/2) = 3 or cotC = 7/rt3. or tanC = (rt3)/7

Now pick up a calculator and find C.