# Thread: Trig triangle/function (ambiguous cases)

1. ## Trig triangle/function (ambiguous cases)

Hey guys, I need help with another problem,

I have posted a picture of the triangle

"For the triangle ABC you are given angle $\beta$ = 30 degrees, and side c = 3."

a. Find side b so that one triangle is formed with side b being the altitude of the triangle.

b. Find side b so that no triangles are formed.

c. Find side b so that two distinct triangles are formed.

d. Find side b so that one triangle is formed but side b is not the altitude of the triangle.

I really need help with these 4 questions so if possible could you guys include work with the answer? Thanks

2. "For the triangle ABC you are given angle $\beta$ = 30 degrees, and side c = 3."

a. Find side b so that one triangle is formed with side b being the altitude of the triangle.

for b to become the altitude it will have to perpendicular to the base BC

so $3sin(30\,^{\circ}) = \frac{3}{2} = b$
recall that $sin30\,^{\circ}$ is $\frac{1}{2}$ so multiply it by 3 and you have $\frac{3}{2}$

ambiguous triangles ussually use the law of sines to determine the possiblities.

3. Could you help me with the other ones as well?

4. hopefully this will help in understanding
how the law of sines works with ambiguous triangles
and too that it will address the answers to you other questions.
see attached image for reference

use the law of sines, to find possible values of $\beta$
if the value of $sin\beta > 1$
there is no solution to this equation and no triangle is possible

if the value of $sin\beta = 1$,
there is one (right) triangle possible

if the value of $sin\beta < 1$,
there are two solutions for $\beta$

$\beta=sin^{-1}\frac{\mathrm{a}sin\mathrm{b}}{\mathrm{a}}\$ and $\ \beta = 180,^{\circ} - sin^{-1}\frac{\mathrm{a}sin\mathrm{b}}{\mathrm{a}}$

if both of solutions, are substituted into
$\alpha + \beta + \gamma$
result as a positive value for $\gamma$,
then two triangles are possilbe
otherwise there is just one triangle
SSA is often an ambiguous case.

5. arg, I can't figure it out! =(

6. the only answer I can see to this is

if b is greater than $\frac{3}{2}$ then you have two triangles

if b is less than $\frac{3}{2}$ there is no triangle

the way the question is worded is kinda difficult to understand how they want an answer because the only lenght that can be calculated is if it is just one triangle

at least that is all I can out of this and am sure they are trying to show how the law of sines is used.

maybe too by now it was explained