How do you solve the triangle given $A=36^circ, a=8, b=5$ ?

Question

7262 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T20:41:14

Please open The Ambiguous Case and read it.

The reference says that if $a > b$ then one possible triangle exists.

Use The Law of Sines to solve for $B$ :

$sin(B)/b = sin(A)/a$

$sin(B) = sin(A)b/a$

$B = sin^-1(sin(A)b/a)$

$B = sin^-1(sin(36^@)5/8)$

$B ~~ 21.55^@$

We know A and B subtract them from 180:

$C ~~ 180 - 36 -21.55$

$C ~~ 122.45^@$

Use The Law of Sines to solve for c:

$c = sin(C)/sin(A)a$

$c = sin(122.45)/sin(36)8$

$c ~~ 11.49$

How do you solve the triangle given A=36^circ, a=8, b=5A=36^circ, a=8, b=5?