Question

In the diagram B, C and D lie on a straight line with $\angle ACD={100}^{\circ },\angle ADB=x^{\circ },\angle ABD=2x^{\circ }and\angle DAC=\angle BAC=y^{\circ }$, then the value of $(\sin y^{\circ }.\tan y^{\circ }+\sec y^{\circ })$ equals

• A. $\frac{7}{2}$
• B. $3$
• C. $\frac{5}{2}$
• D. $5$

Answer 1

The correct option is A $\frac{7}{2}$

In $△ACD$
By Angle sum property,
$\angle ACD+\angle ADC+\angle DAC=180$
$100+x+y=180$
$x+y={80}^{\circ }$..(i)
Also, by exterior angle property,
$\angle BAC+\angle ABC=\angle ACD$
$2x+y=100$..(ii)
Solving (i) and (ii)
$x={20}^{\circ }$, $y={60}^{\circ }$
Now, $\sin y^{\circ }.\tan y^{\circ }+\sec y^{\circ }$
= $\sin 60\tan 60+\sec 60$
= $\left(\frac{\sqrt{3}}{2}\right)\left(\sqrt{3}\right)+2$
= $\frac{3}{2}+2$
= $\frac{7}{2}$