# Math Help - Law of Sines problem (SSA)

1. ## Law of Sines problem (SSA)

The question reads:

Solve triangle ABC if A = 75°, a = 5, and b = 7.
a) B = 43.6°, C = 61.4°, c = 6.4
b) B = 136.4°, C = 31.6°, c = 2.7
c) B = 46.4°, C = 58.6°, c = 4.4
d) no solution

First, I try to solve for angle B:
$
\frac {a} {\sin A} = \frac {b} {\sin B}
$

$
\frac {5} {\sin75} = \frac {7} {\sin B}
$

$
5 \sin B = 7\sin 75
$

$
\sin B = \frac {7sin 75} {5}
$

$
\sin B = 1.3523
$

$
\sin^-1 (1.3523) =$
Undefined

So is the answer "No Solution"? Or, did I make a mistake?

2. Originally Posted by AlderDragon
The question reads:

Solve triangle ABC if A = 75°, a = 5, and b = 7.
a) B = 43.6°, C = 61.4°, c = 6.4
b) B = 136.4°, C = 31.6°, c = 2.7
c) B = 46.4°, C = 58.6°, c = 4.4
d) no solution

First, I try to solve for angle B:
$
\frac {a} {\sin A} = \frac {b} {\sin B}
$

$
\frac {5} {\sin75} = \frac {7} {\sin B}
$

$
5 \sin B = 7\sin 75
$

$
\sin B = \frac {7sin 75} {5}
$

$
\sin B = 1.3523
$

$
\sin^-1 (1.3523) =$
Undefined

So is the answer "No Solution"? Or, did I make a mistake?
I get an answer of no solution too using the sine rule

3. Originally Posted by AlderDragon
The question reads:

Solve triangle ABC if A = 75°, a = 5, and b = 7.
a) B = 43.6°, C = 61.4°, c = 6.4
b) B = 136.4°, C = 31.6°, c = 2.7
c) B = 46.4°, C = 58.6°, c = 4.4
d) no solution
in the ambiguous case, for a triangle to exist, the following inequality must be true ...

$a \ge b\sin{A}$

note that $b\sin{A} = 7\sin(75) = 6.76 > 5$

no triangle is possible

4. Originally Posted by skeeter
in the ambiguous case, for a triangle to exist, the following inequality must be true ...

$a \ge b\sin{A}$

note that $b\sin{A} = 7\sin(75) = 6.76 > 5$

no triangle is possible
Thank you! That clears it up