Question

If A is an acute angle in a right $\Delta ABC,$ right angled at B, then the value of sin A $+$ cos A is:

• A. equal to one
• B. greater than one
• C. less than one
• D. equal to two

Answer 1

The correct option is B greater than one

A is an acute angle of right angle $△ABC$ and right angled at $B.$

$\therefore A+B+C=180°$

$A+90°+C=180°$

$A+C=90°$

Now, $\sin A+\cos A=\sqrt{2}[\frac{1}{\sqrt{2}}\sin A+\frac{1}{\sqrt{2}}\cos A]$

$=\sqrt{2}[\cos 45°\sin A+\sin 45°\cos A]$

$=\sqrt{2}\sin (A+45°)$

When $A=C,$ then $A=C=45°$ $\therefore B=90°$

$\therefore \sqrt{2}\sin (45°+45°)=\sqrt{2}>1$

When $A>C,$ e.g, $A=75°\xi C=15°$

$\Rightarrow \sqrt{2}\sin (75°+45°)=\sqrt{2}\sin 120°=\sqrt{2}\times \frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\sqrt{3}}{\sqrt{2}}>1$

When $A<C,$ e.g, $A=15°\xi C=75°$

$\Rightarrow \sqrt{2}\sin (15°+45°)=\sqrt{2}\sin 60°=\sqrt{2}\times \frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\sqrt{3}}{\sqrt{2}}>1$

Hence, in all cases $\sin A+cosA>1.$

Hence, the answer is greater than one.