Question

o is the centre of the circle is a st is tangent to the circle at d. ∠abo=30°, ∠bds=64° find:
∠bad
∠bcd
∠adt
CORRECT ANSWER WILL BE GIVEN BRAINLIEST
WORST ANSWERS ARE STRICTLY NOT ALLOWED

Answer 1

Answer:

∠BAD=64°

∠BCD=116°

∠ADT=46°

Step-by-step explanation:

• Given ST is the tangent to the circle so ∠ODS=90°
• ∠BAD=64°   [Angles in the alternate segment]
• ABCD is a cyclic quadrilateral and the sum of opposite angles in a cyclic quadrilateral is 180°.
• So,∠BCD+∠BAD=180°

          ⇒∠BCD=180°-64°

          ∴ ∠BCD=116°

• In ΔOBD, ∠ODB=90°-∠BDS

                     ⇒∠ODB=90°-64°

                     ∴∠ODB=26°=∠OBD      [∵OB=OD]

• In ΔABD, ∠ADB=180°-(∠BAD+∠DBA)

                      ⇒∠ADB=180°-(64°+56°)

                      ⇒∠ADB=180°-110°

                      ∴ ∠ADB=70°

• ∴ ∠ODA=∠ADB-∠ODB

                      =70°-26°

                      =44°

• Now,∠ADT+∠ODA=∠ODT=90°   [∵ST is the tangent]

             ⇒∠ADT=90°-44°

             ∴ ∠ADT=46°

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