# Thread: Law of Sines problem (SSA)

1. ## Law of Sines problem (SSA)

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:
$\displaystyle \frac {a} {\sin A} = \frac {b} {\sin B}$

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

$\displaystyle 5 \sin B = 7\sin 75$

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

$\displaystyle \sin B = 1.3523$

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

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

2. Originally Posted by AlderDragon

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:
$\displaystyle \frac {a} {\sin A} = \frac {b} {\sin B}$

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

$\displaystyle 5 \sin B = 7\sin 75$

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

$\displaystyle \sin B = 1.3523$

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

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 ...

$\displaystyle a \ge b\sin{A}$

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

no triangle is possible

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

$\displaystyle a \ge b\sin{A}$

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

no triangle is possible
Thank you! That clears it up