Question

In a \[\Delta ABC\], if \[\angle C = {30^ \circ }\], a = 47 cm and b = 94 cm, then which of the following is true?
A. Right angled
B. Right angled isosceles
C. Isosceles
D. Obtuse angled

Answer 1

Hint: By using cosine law we will find the remaining side of the triangle. Then using the sine law we will find the angles of the given triangle.

Formula used:
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
Sine law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Complete step by step solution:
Given that, \[\angle C = {30^ \circ }\], a = 47 cm and b = 94 cm.
Now using \[{c^2} = {a^2} + {b^2} - 2ab\cos C\] we will find the side c.
Substitute \[\angle C = {30^ \circ }\], a = 47 cm and b = 94 cm in \[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
\[{c^2} = {47^2} + {94^2} - 2 \cdot 47 \cdot 94\cos {30^ \circ }\]
\[ \Rightarrow {c^2} = 2209 + 8836 - 2 \cdot 47 \cdot 94 \cdot \dfrac{{\sqrt 3 }}{2}\]
\[ \Rightarrow {c^2} = 2209 + 8836 - 7652.2004\]
\[ \Rightarrow {c^2} = 3392.7995\]
Taking square roots on both sides
\[ \Rightarrow c \approx 58.248\]cm
Now substitute \[\angle C = {30^ \circ }\], a = 47 cm, b = 94cm, \[c = 58.248\]cm in the sine law
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]:
\[\dfrac{{\sin A}}{{47}} = \dfrac{{\sin B}}{{94}} = \dfrac{{\sin {{30}^ \circ }}}{{58.248}}\]
Therefore,
\[\dfrac{{\sin A}}{{47}} = \dfrac{{\sin {{30}^ \circ }}}{{58.248}}\]
\[ \Rightarrow \sin A = 47 \times \dfrac{{\dfrac{1}{2}}}{{58.248}}\]
\[ \Rightarrow A = {23.79^ \circ }\]or \[{156.21^ \circ }\]
And \[\dfrac{{\sin B}}{{94}} = \dfrac{{\sin {{30}^ \circ }}}{{58.248}}\]
\[ \Rightarrow \sin B = 94 \times \dfrac{{\dfrac{1}{2}}}{{58.248}}\]
\[ \Rightarrow \sin B = 0.80689\]
\[ \Rightarrow \sin B = 53.79\,\,or\,\,\,126.20\]
We know that the sum of all interior angle of a triangle is \[{180^ \circ }\].
If \[\angle A = {23.79^ \circ }\], \[\angle B = {126.20^ \circ }\], then \[\angle A + \angle B + \angle C = {180^ \circ }\]
If \[\angle A = {156.21^ \circ }\], \[\angle B = {53.248^ \circ }\], then \[\angle A + \angle B + \angle C > {180^ \circ }\]
Thus the angles of the triangle are \[\angle A = {23.79^ \circ }\], \[\angle B = {126.20^ \circ }\].
Since \[\angle B = {126.20^ \circ }\] is an obtuse angle, then triangle is an obtuse triangle.
Hence option D is the correct option.

Note: Students often do a common mistake to calculate \[\angle A\] and \[\angle B\]. They do consider the condition \[\sin \left( {\pi - \theta } \right) = \sin \theta \]. So they are unable to get the correct value of \[\angle B\]. The correct value of \[\angle B\] is \[{126.20^ \circ }\].