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 actual angle of right angled $△ABC$ and right angle at $B.$

$\therefore A+B+C={180}^{\circ }$

$A+{90}^{\circ }+C={180}^{\circ }$

$A+C={90}^{\circ }$

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

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

$=\sqrt{2}\sin (A+{45}^{\circ })$

$\left(i\right)$ When $A=C,$ then $A=C={45}^{\circ }$ because $B={90}^{\circ }$

$\Rightarrow \sqrt{2}\sin ({45}^{\circ }+{45}^{\circ })=\sqrt{2}\sin {90}^{\circ }=\sqrt{2}>1$

$\left(ii\right)$ When $A>C,$ e.g. $A={75}^{\circ }\&C={75}^{\circ }$

$\Rightarrow \sqrt{2}\sin ({75}^{\circ }+{45}^{\circ })=\sqrt{2}\sin {120}^{\circ }=\sqrt{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2}}>1$

$\left(iii\right)$ When $A<C,$ e.g, $A={15}^{\circ }\&C={75}^{\circ }$

$\Rightarrow \sqrt{2}\sin ({15}^{\circ }+{45}^{\circ })=\sqrt{2}\sin {60}^{\circ }=\sqrt{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2}}>1$

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

Hence, the answer is greater than one.