Question

In a $$\Delta \,ABC;$$   if $$2\Delta = {a^2} - {\left( {b - c} \right)^2}$$    then value of $$\tan A =$$

A. $$ - \frac{4}{3}$$
B. $$ \frac{4}{3}$$  
C. $$ \frac{8}{15}$$
D. $$ \frac{4}{15}$$
Answer :   $$ \frac{4}{3}$$
Solution :
$$\eqalign{ & 2\Delta = \left( {a - b + c} \right)\left( {a + b - c} \right) \cr & \Rightarrow 2\Delta = 4\left( {s - b} \right)\left( {s - c} \right) \cr & \Rightarrow 2 = \sqrt {\frac{{s\left( {s - a} \right)}}{{\left( {s - b} \right)\left( {s - c} \right)}}} \cr & \Rightarrow \tan \frac{A}{2} = \frac{1}{2} \cr & \therefore \tan A = \frac{{2 \times \frac{1}{2}}}{{1 - \frac{1}{4}}} = \frac{4}{3} \cr} $$

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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