Question

6. Two lines AB and CD intersect at O. If
LAOC = 50°, find LAOD, BOD and BOC​

Answer 1

Answer:

hî your answer ☝️☝️☝️☝️☝️

Answer 2

Given:

• $\angle$ AOC = 50°

Find:

• $\angle$ AOD, $\angle$ BOD angle $\angle$ BOC

Solution:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \thickliness\put(2, 2){\line(2,1){5}} \put(9,2){\line( -4,1){7}} \put(9,2.2){\sf {\large {B}}} \put(2,3.8){ \sf{\large{ A}}} \put(1.8,1.8){ \sf{\large{ C}}} \put(7,4.4){ \sf{\large{ D}}} \put(4.5,3.1){ \sf{\large{ O}}} \put(3.2,3){ \sf{\large{{50}^{ \circ} }} } \put(4.2,3.12){\circle* {0.2}}\end{picture}$

we, know

$: \to \quad \sf \angle AOC + \angle AOD = {180}^{ \circ} \qquad \big \lgroup{ \because Linear \: Pair}\big \rgroup$

where,

• $\angle$ AOC = 50°

So,

$\dashrightarrow\sf \angle AOC + \angle AOD = {180}^{ \circ}$

$\dashrightarrow\sf {50}^{ \circ} + \angle AOD = {180}^{ \circ}$

$\dashrightarrow\sf \angle AOD = {180}^{ \circ} - {50}^{ \circ}$

$\dashrightarrow\sf \angle AOD = {130}^{ \circ}$

$\therefore\sf \angle AOD = {130}^{ \circ}$

•••••••••||||||||•••••••||||||||••••♪••||||

Now,

$: \to \quad \sf \angle AOC = \angle BOD \qquad \big \lgroup{ \because Vertical \: Opposite \: Angle }\big \rgroup$

where,

• $\angle$ AOC = 50°

So,

$\dashrightarrow \sf \angle AOC = \angle BOD$

$\dashrightarrow \sf {50}^{ \circ} = \angle BOD$

$\dashrightarrow \sf \angle BOD = {50}^{ \circ}$

$\therefore \sf \angle BOD = {50}^{ \circ}$

•••••••|||||••••••••||||||••••♪••|♪♪♪•••

we, know

$: \to \quad \sf \angle BOC + \angle BOD = {180}^{ \circ} \qquad \big \lgroup{ \because Linear \: Pair}\big \rgroup$

where,

• $\angle$ BOD = 50°

So,

$\dashrightarrow \sf \angle BOC + \angle BOD = {180}^{ \circ}$

$\dashrightarrow \sf \angle BOC + {50}^{ \circ} = {180}^{ \circ}$

$\dashrightarrow \sf \angle BOC = {180}^{ \circ} - {50}^{ \circ}$

$\dashrightarrow \sf \angle BOC = {130}^{ \circ}$

$\therefore \sf \angle BOC = {130}^{ \circ}$

_______________________________

Hence,

• $\angle$ AOD = 130°
• $\angle$ BOD = 50°
• $\angle$ BOC = 130°