Question

Find the unknown sizes of angles.​

Answer 1

20°

Explanation:

In ∆ DBT,by angle sum property,

• <B+<D+<T = 180°
• 45°+50°+<T = 180°
• <T = 180°-45°-50° = 180°-95° = 85°

In ∆ATQ,<DTB = 85° is an exterior angle

So by exterior angle property,

• ext.< = sum of two opposite interior angles.

So 85°=30°+<TQA

• 85°-30° = <TQA
• <TQA = 55°

Clearly we can see in ∆ TAQ and ∆QBQ,

<TQA and <QQB are vertically opposite,.i.e.,they are equal.

• <TQA = <QQB = 55°

In ∆ QQB,by angle sum property,

• <Q1 +<Q2+<B = 180°
• 45°+55°+<Q2 = 180°
• <Q2 = 180°-100° = 80°

Similarly <Q2 and <RQC makes vertically opposite angles,so they are equal.

• <Q2=<RQC = 80°

In ∆ RBE,by angle sum property,

• <R+<B+<E = 180°
• <ERB = 180°-45°-35° = 180°-80° = 100°

We can see <QRC and <ERB makes linear pair.

Hence their sum must be equal to 180° since they lie on same line.

Hence,<QRC+<ERB = 180°

• 100°+<QRC = 180°
• <QRC = 180°-100° = 80°

In ∆QRC,by angle sum property,

<QRC+<RCQ+<RQC = 180°

• x + 80°+80° = 180°
• x° = 180°-80°-80° = 180°-160° = 20°

Hence the unknown value is 20°.