Question

In a $$\vartriangle ABC,a = 8,b = 10$$     and $$c = 12.$$  Then $$C$$ is equal to

A. $$\frac{A}{2}$$
B. $$2A$$  
C. $$3A$$
D. None of these
Answer :   $$2A$$
Solution :
$$\eqalign{ & \cos C = \frac{{{8^2} + {{10}^2} - {{12}^2}}}{{2 \cdot 8 \cdot 10}} = \frac{1}{8} \cr & \cos A = \frac{{{{10}^2} + {{12}^2} - {8^2}}}{{2 \cdot 10 \cdot 12}} = \frac{3}{4}. \cr & \cos 2A = 2{\cos ^2}A - 1 = 2 \times \frac{9}{{16}} - 1 = \frac{1}{8}. \cr & {\text{So, }}\cos C = \cos 2A \cr & \Rightarrow \,\,C = 2A. \cr} $$

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

A. $$QS = SR$$
B. $$QS : SR = PR : PQ$$
C. $$QS : SR = PQ : PR$$
D. None of these
Releted Question 2

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. $${\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}{\text{metres}}$$
D. none of these
Releted Question 3

In a triangle $$ABC,$$  angle $$A$$ is greater than angle $$B.$$ If the measures of angles $$A$$ and $$B$$ satisfy the equation $$3\sin x - 4{\sin ^3}x - k = 0, 0 < k < 1,$$       then the measure of angle $$C$$ is

A. $$\frac{\pi }{3}$$
B. $$\frac{\pi }{2}$$
C. $$\frac{2\pi }{3}$$
D. $$\frac{5\pi }{6}$$
Releted Question 4

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

A. $$\frac{1}{{\sqrt 6 }}$$
B. $${\frac{1}{3}}$$
C. $$\frac{1}{{\sqrt 3 }}$$
D. $$\sqrt {\frac{2}{3}} $$

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