Question

In the given figure $\alpha and\beta$ are measured in degrees. Which one of the following statements is not correct?

• A. $\beta >\alpha$
• B. $\sec \beta =2$
• C. $\tan 3\alpha =\sqrt{3}$
• D. $\sin (\beta -\alpha )=\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\sin (\beta -\alpha )=\frac{1}{\sqrt{2}}$

In $△ADC$
Sum of angles = 180
$\angle ADC+\angle ACD+\angle CAD=180$
$100+\alpha +\beta =180$
$\alpha +\beta ={80}^{\circ }$ (1)
In $△ABC$
Sum of angles = 180
$\angle ABC+\angle ACB+\angle CAB=180$
$2\alpha +\alpha +2\beta =180$
$3\alpha +2\beta =180$ (2)
Multiply (1) by 3 and subtract from (2)
$3\alpha +3\beta -3\alpha -2\beta =240-180$
$\beta ={60}^{\circ }$
And from (1)
$\alpha =80-60={20}^{\circ }$
Thus, $\beta >\alpha$ is true.
$sec\beta =sec60=\sqrt{2}$ is true.
$\tan 3\alpha =\tan 60=\sqrt{3}$ is true.

Hence, $\sin (\beta -\alpha )=\sin 40\ne \frac{1}{\sqrt{2}}$ is not true.