The sides of a triangle are $${\sin \alpha , \cos \alpha }$$ and $$\sqrt {1 + \sin \alpha \cos \alpha } $$ for some $$0 < \alpha < \frac{\pi }{2}.$$ Then the greatest angle of the triangle is
A.
150°
B.
90°
C.
120°
D.
60°
Answer :
120°
Solution :
Let $$a = \sin \alpha ,b = \cos \alpha \,\,{\text{and }}c = \sqrt {1 + \sin \alpha \cos \alpha } $$
Clearly $$a$$ and $$b < 1$$ but $$c > 1$$ as $$\sin \alpha > 0\,\,{\text{and}}\,\,\cos \alpha > 0$$
∴ $$c$$ is the greatest side and greatest angle is $$C$$
$$\eqalign{
& \therefore \,\,\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr
& = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha }}{{2\sin \alpha \cos \alpha }} = - \frac{1}{2} \cr
& \therefore \,\,C = {120^ \circ } \cr} $$
Releted MCQ Question on Trigonometry >> Properties and Solutons of Triangle
Releted Question 1
If the bisector of the angle $$P$$ of a triangle $$PQR$$ meets $$QR$$ in $$S,$$ then
From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is 15°. The distance of the boat from the foot of the light house is
A.
$$\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)60\,{\text{metres}}$$
B.
$$\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)60\,{\text{metres}}$$
C.
$${\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}{\text{metres}}$$
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In a triangle $$ABC,$$ $$\angle B = \frac{\pi }{3}{\text{ and }}\angle C = \frac{\pi }{4}.$$ Let $$D$$ divide $$BC$$ internally in the ratio 1 : 3 then $$\frac{{\sin \angle BAD}}{{\sin \angle CAD}}$$ is equal to