In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

Concept:

Consider a triangle ABC,

Cosine rule:

\(\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}\)

Sine rule:

\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}\)

Calculation:

Given: a = (1 + √ 3) cm, b = 2 cm and ∠C = 60°

We know that, \(\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}\)

\( \Rightarrow \cos 60^\circ = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2} + {2^2} - {{\rm{c}}^2}}}{{2\left( {1 + \sqrt 3 } \right)\left( 2 \right)}}\)

⇒ 2(1 + √3) = 1 + 3 + 2√3 + 4 – c²

⇒ c² = 6

⇒ c = √6

Using sin rule we get,

\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}\)

\( \Rightarrow \frac{{1 + \sqrt 3 }}{{\sin {\rm{A}}}} = \frac{2}{{\sin {\rm{B}}}} = \frac{{\sqrt 6 }}{{\sin 60^\circ }}\)

⇒ sin B = (2 / √6) × sin 60°

\( \Rightarrow \sin {\rm{B}} = \frac{1}{{\sqrt 2 }}\)

⇒ B = 45°

Now, sum of all angles of triangles = 180°

⇒ A + B + C = 180°

⇒ A = 180° – 45° – 60°

⇒ A = 75°