Question

The greater of two angles A = 2 tan^–1 ($2 \sqrt { 2 }$–1) and

B = 3 sin^–1 (1/3) + sin^–1 (3/5) is

• A. A
• B. B
• C. Both are equal
• D. None of these

Answer 1

Answer: (A)

A

Answer 2

A = 2 tan^–1 $( 2 \sqrt { 2 } - 1 )$

= 2 tan^–1 (2 × 1.414 –1) = 2 tan^–1 (1.828)

Ž A > 2 tan^–1 $\sqrt { 3 }$ = 2p/3

For angle B, sin^–1$\frac { 1 } { 3 }$< sin^–1 $\frac { 1 } { 2 }$ = $\frac { \pi } { 6 }$ Ž 3 sin^–1 $\left( \frac { 1 } { 3 } \right)$ < $\frac { \pi } { 2 }$

3 sin^–1$\left( \frac { 23 } { 27 } \right)$

= sin^–1 (.851) < sin^–1$\frac { \sqrt { 3 } } { 2 }$=$\frac { \pi } { 3 }$ Ž 3 sin^–1 $\left( \frac { 1 } { 3 } \right)$ <$\frac { \pi } { 3 }$

Also sin^–1 $\frac { \pi } { 3 }$

so sin^–1$\frac { \pi } { 3 }$

Adding B = 3 sin^–1$\frac { 1 } { 3 }$+ sin^–1 $\frac { 3 } { 5 }$< $\frac { 2 \pi } { 3 }$ So A > B