Question

In the figure shown below, $m\angle ABD=2x,m\angle ADB=3x,m\angle APB=4xandm\angle CBQ=7x$. Therefore, $m\angle BCD$ is

• A. ${90}^{\circ }$
• B. ${85}^{\circ }$
• C. ${72}^{\circ }$
• D. ${96}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

Refer Figure

Given :

$\angle ABD=2x$

$\angle ADB=3x$

$\angle APB=4x$

$\angle CBQ=7x$

Solution:

$\angle ABD=\angle ACD=2x$ (Angles subtended by a chord in the same segment)

$\angle ADB=\angle ACB=3x$ (Angles subtended by a chord in the same segment)

$\angle APB=\angle DPC=4x$ (Opposite angles)

In straight angle, $ABQ$,

$\angle PBC=180°-\angle ABD-\angle QBC$

$\angle PBC=180°-2x-7x=180°-9x$

In $△CPD$

$\angle CDP=180°-\angle CPD-\angle ACD$

$\angle CDP=180°-4x-2x=180°-6x$

In quad ABCD

$\angle ABC+\angle ADC=180°$ (sum of opposite angles of a cycli quadrilateral)

$2x+180°-9x+3x+180°-6x=180°$

$10x=180°$

$x=18°$

$\angle BCD=\angle BCA+\angle DCA=3x+2x=5x=5\times 18°=90°$

$\angle BCD=90°$