Answer:
One triangle
Step-by-step explanation:
Find the measures of the triangle ABC
Let
D ----> intersection point of the height triangle ABC with the segment AB
step 1
Find the measure of angle B
In the right triangle BCD
[tex]sin(B)=\frac{CD}{BC}[/tex] ---> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(B)=\frac{9}{15}[/tex]
using a calculator
[tex]B=sin^{-1}(\frac{9}{15})=36.9^o[/tex]
step 2
Find the length of segment DB in the right triangle BCD
Applying the Pythagorean Theorem
[tex]BC^2=DC^2+DB^2[/tex]
substitute
[tex]15^2=9^2+DB^2[/tex]
[tex]DB^2=15^2-9^2[/tex]
[tex]DB^2=144\\DB=12\ in[/tex]
step 3
Find the measure of angle C
we know that
The sum of the interior angles in any triangle must be equal to 180 degrees
so
[tex]A+B+C=180^o[/tex]
substitute the given values
[tex]33^o+36.9^o+C=180^o[/tex]
[tex]C=180^o-69.9^o=110.1^o[/tex]
step 4
Find the length side AC
In the right triangle ACD
[tex]sin(A)=\frac{CD}{AC}[/tex] ---> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(33^o)=\frac{9}{AC}[/tex]
solve for AC
[tex]AC=\frac{9}{sin(33^o)}=16.5\ in[/tex]
step 5
Find the length of segment AD
In the right triangle ACD
[tex]tan(A)=\frac{CD}{AD}[/tex] ---> by TOA (opposite side divided by the adjacent side)
substitute the given values
[tex]tan(33^o)=\frac{9}{AD}[/tex]
[tex]AD=\frac{9}{tan(33^o}}=13.9\ in[/tex]
step 6
Find the length side AB
[tex]AB=AD+DB[/tex]
substitute the given values
[tex]AB=13.9+12=25.9\ in[/tex]
step 7
The measures of triangle ABC are
[tex]A=33^o\\B=36.9^o\\C=110.1^o\\AC=16.5\ in\\CB=15\ in\\AB=25.9\ in[/tex]
so
The measurements of the ABC triangle are unique
therefore
Only one triangle can be constructed with the given measurements
Two distinct triangles are possible with the given measurements.
Let the sides AC and AB are the fixed sides and BC is a swinging
side of ΔACB.
a). If CB < 9 (Height of the triangle),
No triangle is possible.
b). If CB = 9,
Only one triangle is possible.
c). If CB > 9 and CB > AC,
Only one triangle is possible.
d). If CB > 9 and CB < AC,
Two triangles are possible.
Find the length of the fixed line AC from the given ΔACB,
sin(33°) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
= [tex]\frac{9}{AC}[/tex]
AC = [tex]\frac{9}{\text{sin}(33)^\circ}[/tex]
AC = 16.525
AC ≈ 16.53 in.
Since, CB < AC and CB > 9 (height of the triangle),
Therefore, two triangles are possible with the given measurements.
Learn more about the properties of the triangles here,
https://brainly.com/question/4221200?referrer=searchResults
