Question

If O is the center of the circle, find the value of x in each of the following figures (using the given information):​

Answer 1

$\huge\boxed{\fcolorbox{red}{pink}{Solution}}$

(i) $∠ACB=∠ADB$ (Angles in the same segment of a circle)

But $∠ADB=x°$

$⇒∠ABC=x°$

Now in $ΔABC$

$∠CAB+∠ABC+∠ACB=180°$

$⇒40°+90°+x° =180°$ (AC is the diameter)

$⇒130°+x °=180°$

$⇒x°=180°−130°=50°$

(ii) $∠ACD=∠ABD$ (angles in the same segment)

And, $∠ACD=x°$

Now in triangle OAC,

OA=OC (radii of the same circle)

$⇒∠ACO=∠AOC$ (opposite angles of equal sides)

Therefore,$x° =62°$

(iii)$∠AOB+∠AOC+∠BOC=360°$

(sum of angles at a point)

$⇒∠AOB+80°+130°=360°$

$⇒∠AOB+210°=360°$

$⇒∠AOB=360°−210°=150°$

Now arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

$∴∠AOB=2∠ACB$

$⇒∠ACB= \frac{2}{1}∠AOB= \frac{2}{1}×150 °=75°$

(iv)$∠ACB+∠CBD=180°$

$⇒∠ABC+75°=180°$

$⇒∠ABC=180° −75 °=105°$

Now arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle.

Relfex$∠AOC=2∠ABC</p><p>[tex]=2×105°$

$=210°$

(v) $∠AOC+∠COB=180°$

$➪135° +∠COB=180°$

$➪∠COB=180°−135°=45°$

Now arc BC Subtends reflex ∠COB at the centre and ∠CDB at the remaining part of the circle.

$➪∠COB=2∠CDB$

$➪∠CDB=\frac{2}{1} ∠COB= \frac{2}{1}×45°= \frac{45°}{2} =22\frac{2}{1}°$

(vi) Arc AB subtends ∠AOD at the centre and ∠ACD at the remaining part of the Circle

∠AOD=2∠ACB

$➪∠ACB= \frac{2}{1}∠AOD= \frac{2}{1}×70°=35 °$

$➪∠CMO=90°$

$➪∠AMC=90°$

$(∠AMC+∠CMO=180° )$

Now in ΔACM

$∠ACM+∠AMC+∠CAM=180°$

$➪35°+90° +x °=180°$

$➪125°+x°=180°$

$➪x° =180−125° =55°$

Answer 2

Step-by-step explanation:

(i) ∠ACB=∠ADB∠ACB=∠ADB (Angles in the same segment of a circle)

But ∠ADB=x°∠ADB=x°

⇒∠ABC=x°⇒∠ABC=x°

Now in ΔABCΔABC

∠CAB+∠ABC+∠ACB=180°∠CAB+∠ABC+∠ACB=180°

⇒40°+90°+x° =180°⇒40°+90°+x°=180° (AC is the diameter)

⇒130°+x °=180°⇒130°+x°=180°

⇒x°=180°−130°=50°⇒x°=180°−130°=50°

(ii) ∠ACD=∠ABD∠ACD=∠ABD (angles in the same segment)

And, ∠ACD=x°∠ACD=x°

Now in triangle OAC,

OA=OC (radii of the same circle)

⇒∠ACO=∠AOC⇒∠ACO=∠AOC (opposite angles of equal sides)

Therefore,x° =62°x°=62°

(iii)∠AOB+∠AOC+∠BOC=360°∠AOB+∠AOC+∠BOC=360°

(sum of angles at a point)

⇒∠AOB+80°+130°=360°⇒∠AOB+80°+130°=360°

⇒∠AOB+210°=360°⇒∠AOB+210°=360°

⇒∠AOB=360°−210°=150°⇒∠AOB=360°−210°=150°

Now arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴∠AOB=2∠ACB∴∠AOB=2∠ACB

⇒∠ACB= \frac{2}{1}∠AOB= \frac{2}{1}×150 °=75°⇒∠ACB=

1

2

∠AOB=

1

2

×150°=75°

(iv)∠ACB+∠CBD=180°∠ACB+∠CBD=180°

⇒∠ABC+75°=180°⇒∠ABC+75°=180°

⇒∠ABC=180° −75 °=105°⇒∠ABC=180°−75°=105°

Now arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle.

Relfex∠AOC=2∠ABC < /p > < p > [tex]=2×105°∠AOC=2∠ABC</p><p>[tex]=2×105°

=210°=210°

(v) ∠AOC+∠COB=180°∠AOC+∠COB=180°

➪135° +∠COB=180°➪135°+∠COB=180°

➪∠COB=180°−135°=45°➪∠COB=180°−135°=45°

Now arc BC Subtends reflex ∠COB at the centre and ∠CDB at the remaining part of the circle.

➪∠COB=2∠CDB➪∠COB=2∠CDB

➪∠CDB=\frac{2}{1} ∠COB= \frac{2}{1}×45°= \frac{45°}{2} =22\frac{2}{1}°➪∠CDB=

1

2

∠COB=

1

2

×45°=

2

45°

=22

1

2

°

(vi) Arc AB subtends ∠AOD at the centre and ∠ACD at the remaining part of the Circle

∠AOD=2∠ACB

➪∠ACB= \frac{2}{1}∠AOD= \frac{2}{1}×70°=35 °➪∠ACB=

1

2

∠AOD=

1

2

×70°=35°

➪∠CMO=90°➪∠CMO=90°

➪∠AMC=90°➪∠AMC=90°

(∠AMC+∠CMO=180° )(∠AMC+∠CMO=180°)

Now in ΔACM

∠ACM+∠AMC+∠CAM=180°∠ACM+∠AMC+∠CAM=180°

➪35°+90° +x °=180°➪35°+90°+x°=180°

➪125°+x°=180°➪125°+x°=180°

➪x° =180−125° =55°➪x°=180−125°=55°