Question

In a triangle ABC, tan = $\frac { 5 } { 6 }$, tan $\frac { \mathrm { C } } { 2 }$ = $\frac { 2 } { 5 }$, then –

• A. a, c, b are in A.P.
• B. a, b, c are in A.P.
• C. b, a, c are in A.P.
• D. a, b, c are in G.P.

Answer 1

Answer: (B)

a, b, c are in A.P.

Answer 2

tan= $\frac { 5 } { 6 }$ , tan $\frac { \mathrm { C } } { 2 }$ = $\frac { 2 } { 5 }$

$\sqrt { \frac { ( \mathrm { s } - \mathrm { b } ) ( \mathrm { s } - \mathrm { c } ) } { \mathrm { s } ( \mathrm { s } - \mathrm { a } ) } }$ = $\frac { 5 } { 6 }$ …….. (1)

$\sqrt { \frac { ( \mathrm { s } - \mathrm { a } ) ( \mathrm { s } - \mathrm { b } ) } { \mathrm { s } ( \mathrm { s } - \mathrm { c } ) } }$ =$\frac { 2 } { 5 }$ …….. (2)

(1) × (2)

= $\frac { 5 } { 6 }$×$\frac { 2 } { 5 }$

×$\left( \frac { 2 } { a + b + c } \right)$ = $\frac { 1 } { 3 }$

̃ 3a – 3b + 3c = a + b + c

̃ 2a + 2c = 4b

̃ a + c = 2b

̃ a + c = 2b

̃ a, b, c are in A.P.