If you consider
Option (A) in which
b = 14, c = 6, a = 14.9→→→→Sides of Triangle
A= 52°, C = 24.2°, B = 103.8°→→Angles of Triangle
As , you can see that sum of two sides is greater than third side
And , sum of three angles of triangle is 180°.
b +c= 14 +6=20>a=14.9
c+a=6+14.9=20.9>b=14
a+b=14.9 +14=28.9>c=6
Sum of the angles = (52)° + (24.2)°+ (103.8)°
= (52)° + (128)°
= 180°
So, yes ,triangle is Possible.
Answer:
B≈103.8° , a≈11.3 and C≈24.2°
B is correct.
Step-by-step explanation:
Given: [tex]A=52^\circ, b=14 \text{ and } c=6[/tex]
Using sine find rest part of the triangle.
Sine law:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
Cosine Law:
[tex]a^2=b^2+c^2-2bc\cos A[/tex]
where, [tex]A=52^\circ, b=14 \text{ and } c=6[/tex]
[tex]\dfrac{a}{\sin 52^\circ}=\dfrac{14}{\sin B}=\dfrac{6}{\sin C}[/tex]
[tex]a^2=14^2+6^2-2(14)(6)\cos 52^\circ=158.57[/tex]
[tex]a\approx 11.3[/tex]
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}[/tex]
[tex]\dfrac{11.3}{\sin 52^\circ}=\dfrac{6}{\sin B}[/tex]
[tex]\sin B=0.4184[/tex]
[tex]B\approx 103.8[/tex]
[tex]A+B+C=180^\circ[/tex]
[tex]C\approx 180-52-103.8 = 24.2[/tex]
Hence, B≈103.8° , a≈11.3 and C≈24.2°
