How do you determine the number of possible triangles and find the measure of the three angles given #a=9, c=10, mangleC=150#?

Question

Asher Friedman · Accepted Answer

#A=26^@45', B=150^@, C=3^@15'#

Since the given information is for a SSA triangle it is the ambiguous case. In the ambiguous case we first find the height by using the formula #h=bsin A#.

Note that A is the given angle and its side is always a so the other side will be b .

So if #A < 90^@# and if

#h < a < b# then then there are two solutions or two triangles.

#h < b < a# then there is one solution or one triangle.

#a < h < b# then there is no solution or no triangle.

If #A >=90^@# and if

#a > b# then there is one solution or one triangle.

#a <=b# there is no solution

#h=9 sin150^@=4.5#, since #4.5 < 9 < 10 # we have

#h < b < a# so we are looking for one solution. Hence,

#Sin A/a = sin B / b#

#sin A /9 = sin 150^@/10#

#sin A =(9 sin 150^@)/10#

#A=sin^-1 ((9 sin 150^@)/10)=26^@45'#

and therefore

#C=180^@-150^@-26^@ 45'=3^@15'#

Kai Anderson · Answer

undefined To determine the number of possible triangles and find the measures of the three angles given side lengths $ a = 9 $, $ c = 10 $, and $ \angle C = 150^\circ $, we can use the Law of Sines and the properties of triangles.

First, we use the Law of Sines to find the length of side $ b $:

$$ \frac{a}{\sin A} = \frac{c}{\sin C} $$

Solving for $ \sin A $:

$$ \sin A = \frac{a \cdot \sin C}{c} $$

$$ \sin A = \frac{9 \cdot \sin 150^\circ}{10} $$

$$ \sin A = \frac{9 \cdot \frac{\sqrt{3}}{2}}{10} $$

$$ \sin A = \frac{9\sqrt{3}}{20} $$

$$ \sin A \approx 0.389 $$

Since $ A $ must be acute, there is only one possible triangle.

Now, we can find $ B $ using the fact that the sum of the angles in a triangle is $ 180^\circ $:

$$ A + B + C = 180^\circ $$

$$ B = 180^\circ - A - C $$

$$ B = 180^\circ - \arcsin(0.389) - 150^\circ $$

$$ B \approx 40.6^\circ $$

Finally, we find $ \angle A $ using the fact that the sum of the angles in a triangle is $ 180^\circ $:

$$ A = 180^\circ - B - C $$

$$ A = 180^\circ - 40.6^\circ - 150^\circ $$

$$ A \approx -10.6^\circ $$

However, $ A $ cannot be negative, so we have made a mistake. Let's recompute $ A $:

$$ A = 180^\circ - B - C $$

$$ A = 180^\circ - 40.6^\circ - 150^\circ $$

$$ A = -10.6^\circ $$

This result indicates that our assumption that $ A $ is acute is incorrect. Therefore, there is no triangle with the given side lengths and angle measures.