Question

If $A=130°$ and $x=\sin A+\cos A$, then,

• A. $x>0$
• B. $x<0$
• C. $x=0$
• D. $x\le 0$

Answer 1

The correct option is A

$x>0$

Explanation for correct option:

Finding the value:

Given, $A=130°$

Now,

$\begin{array}{rcl}x& =& \sin A+\cos A\\ x& =& \sin 130°+\cos 130°\\ x& =& \sin (180°–50°)+\cos (180°–50°)\\ x& =& \sin 50°–\cos 50°\\ x& =& \sin 50°–\cos (90°–40°)\\ x& =& \sin 50°–\sin 40°....\left(1\right)\\ & & \end{array}$

In equation

$\begin{array}{rcl}x& =& 2\sin \frac{(50°–40°)}{2}\cos \frac{(50°+40°)}{2}\\ x& =& 2\sin 5°\cos 45°\\ & & \end{array}$$\mathit{s}\mathit{i}\mathit{n}\mathit{C}\mathbf{–}\mathit{s}\mathit{i}\mathit{n}\mathit{D}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{(}\mathbf{C}\mathbf{–}\mathbf{D}\mathbf{)}}{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{(}\mathbf{C}\mathbf{+}\mathbf{D}\mathbf{)}}{\mathbf{2}}$

As we know that $\sin 5°\&\cos 45°$ are positive,

Therefore, $x$ is also positive & varies from $x>0$

Hence, correct option is $\left(A\right)$