Answer:
[tex]\theta_{CAB}=128.316[/tex]
[tex]\theta_{ABC}=25.842[/tex]
[tex]\theta_{BCA}=25.842[/tex]
Step-by-step explanation:
A = (-3,0) , B = (1,3) , and C = (1,-3)
We're going to use the distance formula to find the length of the sides:
[tex]r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}[/tex]
[tex]AB= \sqrt{(-3-1)^2+(0-3)^2}=5[/tex]
[tex]BC= \sqrt{(1-1)^2+(3-(-3))^2}=9[/tex]
[tex]CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5[/tex]
we can use the cosine law to find the angle:
it is to be noted that:
the angle CAB is opposite to the BC.
the angle ABC is opposite to the AC.
the angle BCA is opposite to the AB.
to find the CAB, we'll use:
[tex]BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}[/tex]
[tex]\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}[/tex]
[tex]\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)} [/tex]
[tex]\theta_{CAB}=\arccos{-\dfrac{0.62}} [/tex]
[tex]\theta_{CAB}=128.316[/tex]
Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!
To find the angle ABC
[tex]\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}[/tex]
[tex]\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)[/tex]
[tex]\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)[/tex]
[tex]\theta_{ABC}=\arcsin{0.4359}\right)[/tex]
[tex]\theta_{ABC}=25.842[/tex]
finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.
[tex]\theta_{BCA}=25.842[/tex]
this can also be checked using the fact the sum of all angles inside a triangle is 180
[tex]\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180[/tex]
[tex]25.842+128.316+25.842[/tex]
[tex]180[/tex]
Answer:
The measure of ∠ABC is 53.130º
The measure of ∠BCA is: 53.130º
The measure of ∠CAB is: 73.739º
Step-by-step explanation:
1) Calculate the measure of each leg:
[tex]d_{AB}=\sqrt{(-3-1)^{2}+(0-3)^{2}}\Rightarrow d_{AB}=5\\d_{AC}=\sqrt{(-3-1)^{2}+(0-(-3))^{2}}\Rightarrow d_{AC}=5\\d_{BC}=\sqrt{(1-1)^{2}+(3-(-3)^{2}}\Rightarrow d_{BC}=6\\[/tex]
2) Since it's an isosceles triangle, trace a line segment to turn into two right triangles, from point A to the intersection of BC with x-axis.
From point -3 to point -1
[tex]|-3-1|=3+1=4[/tex]
The height is equal to 4
Then we can use trigonometric functions to calculate the angle measures .The measure of ∠ABC :
Since D is the midpoint, BC was equally sectioned into two parts.
[tex]sen\angle B=\frac{4}{5}\Rightarrow \angle B=arc \:sen\left ( \frac{4}{5} \right )\Rightarrow \angle B=53.13^{\circ}\\\angle B=\angle C=53.130^{\circ} \:isosceles \:triangle\\tan\angle A'=\frac{3}{4}\:\:A'=arc tan \frac{3}{4} \therefore\angle A'=36.87^{\circ}\Rightarrow A'=A''\Rightarrow A'+A''=A\Rightarrow A\approx 73.739^{\circ}[/tex]
