Question 1035792: I have to solve a triangle. Given 2 sides and 1 angle from it. I gotta determine if it's based on 1 triangle, 2 triangles, or no triangles.
a= 19, b= 10, B= 10 degrees.
Answer by Edwin McCravy(20081) (Show Source):
You can put this solution on YOUR website! I have to solve a triangle. Given 2 sides and 1 angle from
it. I gotta determine if it's based on 1 triangle,
2 triangles, or no triangles. a= 19, b= 10, B= 10 degrees.
To get a better understanding, let's sketch the geometry of the
situation. I start by drawing an arbitrarily long line labeling
the endpoint as the vertex of the given angle, in this case B.
At point B we draw the angle B = 10° by drawing side
BC = a = 19:
Next we take a compass and open it to the length
b = CA = 10. Put the sharp point on C and swing
an arc that cuts the horizontal line that we drew
first in two places:
So it appears in this case there are two choices for point A.
Either this way with the green line:
or this way with the blue line:
Now let's do the calculating using the law of sines:
substitute a=19, b=10, B=10°
We use only the first and second parts:
Cross-multiply
Divide both sides by 10
Do the calculation:
[If this had come out greater than 1 there would have
been 0 solutions. If this had come out equal to 1
there would have been 1 solution, and it would have
been a right triangle with hypotenuse b.]
Since it came out less than 1, we can tell there is
either 1 or 2 solutions.
First solution:
Next we calculate the first quadrant (acute) solution
for angle A using inverse sine feature on the calculator
and get angle A = 19.26462015°. Then we can calculate the
third angle C by adding 19.26462015° and 10° and subtracting
from 180°, getting angle C = 150.7353799°.
Now we use the law of sines again:
This time we use only the second and third parts,
and substitute 150.7353799° for angle C
Cross-multiply:
Divide both sides by sin(10°)
-----------------
Next we see if there is a second solution.
We calculate the second quadrant (obtuse) solution
for angle A by finding the complement of 19.26462015°,
by subtracting it from 180°, getting 160.7353798°.
Then we can calculate the third angle C by adding
160.7353798° and 10° and subtracting
from 180°, getting angle C = 9.2646202°.
[If this had come out negative there would have
been only the 1 solution.]
So we have a second solution, and we use the law of
sines to find c just as we did to find the first
solution:
Cross-multiply:
Divide both sides by sin(10°)
Edwin
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