Question

5. In the adjoining figure, ZACB - 40and ZBED - 120°
Find the measure of CBD.​

Answer 1

explain my slon please

Answer 2

Answer:

$20^{\circ}$

Step-by-step explanation:

$\setlength{\unitlength}{1.5cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.1,-2.1)(-0,-2.3)\qbezier(-1.5,1.8)(-1.5,1.8)(-1.7,-1.7)\qbezier(-1.7,-1.65)(-1.7,-1.65)(0.5,2.2)\qbezier(0.5,2.2)(0.5,2.2)(1.2,-1.9)\qbezier(1.2,-1.9)(1.2,-1.9)(-1.5,1.8)\put(-0.55,-0.3){\bf E}\qbezier(-0.3,0.7)(-0.1,0.5)(-0.3,0.2)\put(0,0.4){ \bf 120^{\circ} }\qbezier(-1.5,1.3)(-1.3,1.1)(-1.2,1.3)\put(-1.35,0.8){ \bf 40^{\circ} }\put(-2.1,-2){\bf A}\put(-2,2){\bf C}\put(1.4,-2.4){\bf B}\put(0.5,2.5){\bf D}\qbezier(-1.7,-1.65)(-1.7,-1.65)(1.2,-1.9)\end{picture}$

Construction:-

Join A and B

Solution:-

When we have joined A and B then ACDB is a major segment

We know that angle in the same segment are equal,

Using the above theorem,

Therefore,

$\rm \angle{ACD} = \angle{ADB}$

As in the question, it is given that $\rm \angle ACD = 40^{\circ}$

Therefore,

$\rm \angle{ADB} = 40^{\circ}$

In $\rm \Delta BED$,

$\rm \angle{BED} = 120^{\circ}, \angle{EDB} = \angle{ADB} = 40^{\circ}$

According to angle sum property,

$\rm \angle{EBD} + \angle{BED} + \angle{EDB} = 180^{\circ}$

$\rm \angle{EBD} + 120^{\circ} + 40^{\circ} = 180^{\circ} \implies \angle{EBD} = 20^{\circ}$

As $\rm \angle{CBD} = \angle{EBD}$

Therefore,

$\rm\angle{CBD} = 20^{\circ}$.