I can easily find the measurements of sides with two angles and one side. What i'm having trouble on is completing and finding all values of a triangle based on one angle and two sides. Here I've drawn up a problem for reference.
Help
i trust your want to find out what x is,
you need the sine rule which is
a/sin A = b/sin B = c/sin C
we have numbers for c, sin C, b and sin B (its x) so as b/sin B = c/sin C you just rearrange the algebra to get sin B, then it should be easy from their but i will show you anyway
80.2/sin 67 =56.3/sin x
80.2sin x = 56.3sin 67
sin x = 56.3 sin 67/80.2
x = inverse sin (56.3sin67/80.2)
x = 40.3 degrees.
Hello, Jonathan!
It seems to be straight-forward.
. . Exactly where is your difficulty?
Code:C * * * * 67° * a b=56.3 * * * * * * A * * * * * * * B c=80.2
From the Law of Sines, we have: .
Then: .
. . Hence: .
Then: .
From the Law of Sines, we have: .
Then: .
. . Hence: .
Hey my difficulty lies in how you got
40.25 for angle B.
I know you got .646xxxxx when you found the SIN B figuring it out algebraically. But from this irrational number I don't understand how you got 40.25.
I experimented and used inverse SIN of .646xxxxx and it gave me 40.25. I still don't understand why we use the inverse. This is my pain and difficulty
we use the inverse sin becasue it reverses sin, thats simply why it cancels cout the effect that sin has on the x leaving us with only x. Its the same i idea as knowing that 7x = 14 we use division to reverse the miultiplication, trig is just the same (not very good at exaplaing maths??? hope it helped if not just ask again)