Solve triangle in which is
$$ b+c=20,\ a=5\sqrt 2,\ \gamma=135°$$
($2$ solutions)
u = alpha, v = beta, w = gamma
w = 135, v = 180 - 135 - u = 45 - u
5sqrt(2) / SIN(u) = (20-b) / SIN(135) = b / SIN(45 - u)
Above is a hint.
Sorry, homework not done here...
Not sure what you're solving for...so solve for whatever turns you on!!
Solve triangle means find basic elements of triangle
$$ a,\ b,\ c,\ \alpha,\ \beta,\ \gamma.$$
I have solutions which demand picture which I don't know how to draw. And, please somebody to tell me where to post problems like challenges.
If $D-C-A,\ and\ BD=h_b$ then $BD=DC=5$ and from $\Delta ABD$ with Pithagoras law we have
$$(20-b)^2=5^2+(b+5)^2$$
which gives $b=7$. Te rest is easy, but this give me just one solution.
It's always interesting when the first problem is trying to decipher what the poster is really trying to do. I have attached picture below (not necessarily to scale) of my best guess of what the professor is looking for. I'm guessing the $135^\circ$ may be interior or exterior to the triangle giving an alternate angle $A'$ and corresponding adjacent sides $b'$ and $c'$ to give two triangles.
Here's a picture (click on it to expand):
Walagaster's draw is Ok, same I used, without point A'. $h_b$ is altitude of triangle, and $\Delta BDC$ is isosceles, right triangle. My native language is serbian, and we use specific termilology for elements of triangle for which I'm not sure I know the english terms. I find this problem in book of probems, and author claims that exist 2 solutions. I see only one.
$\alpha=\arcsin\frac {5}{13}$ and $\beta=\frac {7}{13\sqrt 2}$.
$\gamma=135°$ is given interior angle of triangle, so A' is of no use.
BTW, I'm in retirement, and I was not happy with the students. I feel like a sigh whose song nobody heard. Some like you.