Answer:
∠ADB=35°
Explanation:
Since there is a diagonal AC passing through the rhombus ABCD:
∠ACB = ∠ACD (AC is the diagonal and bisector of ∠BCD)
Since, ∠ACB = 55°;
∠ACD = 55° (As shown above:)
∴ ∠BCD = ∠ACB + ∠ACD
= 55° + 55° = 110°
∠BCD = 110°
Now, we know that a rhombus is a parallelogram
∴ opposite angles are supplementary
So, ∠BCD + ∠ADC = 180°
110° + ∠ADC = 180°
∠ADC = 180° - 110°
∠ADC = 70°
Now,
We need to find ∠ADB:
We need to half ∠ADC (Since BC would bisect ∠ABC)
∴ ∠ADB = \frac{1}{2}21 ∠ADC
∠ADB = \frac{1}{2}21 70°
∠ADB = 35°
Final answer: ∠ADB = 35°
Hope it helps,
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Answers & Comments
Answer:
∠ADB=35°
Explanation:
Since there is a diagonal AC passing through the rhombus ABCD:
∠ACB = ∠ACD (AC is the diagonal and bisector of ∠BCD)
Since, ∠ACB = 55°;
∠ACD = 55° (As shown above:)
∴ ∠BCD = ∠ACB + ∠ACD
= 55° + 55° = 110°
∠BCD = 110°
Now, we know that a rhombus is a parallelogram
∴ opposite angles are supplementary
So, ∠BCD + ∠ADC = 180°
110° + ∠ADC = 180°
∠ADC = 180° - 110°
∠ADC = 70°
Now,
We need to find ∠ADB:
We need to half ∠ADC (Since BC would bisect ∠ABC)
∴ ∠ADB = \frac{1}{2}21 ∠ADC
∠ADB = \frac{1}{2}21 70°
∠ADB = 35°
Final answer: ∠ADB = 35°
Hope it helps,