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Question
The angles of a triangle formed by 2 adjacent sides and a diagonal of a parallelogram are in the ratio 1 : 5 : 3. Calculate the measures of all the angles of the parallelogram.
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Solution

ABCD is a parallelogram.
Let ∠CAB = x°
Then, ∠ABC = 5x° and ∠BCA = 3x°
In ΔABC,
∠CAB + ∠ABC + ∠BCA = 180° ...(sum of angles of triangle = 180°)
x° + 5x° + 3x° = 180°
9x° = 180°
x° = 20°
⇒ ∠CAB = x° = 20°
⇒ ∠ABC = 5x° = 5 x 20° = 100°
⇒ ∠ BCA = 3x° = 3 x 20° = 60°
Now,
∠ADC = ∠ABC = 100° ...(opposite angles of a parallelogram are equal)
∠ACD =∠CAB = 20° ...(Alternate angles since BC || AD)
∠ CAD = ∠BCA = 60° ...(Alternate angles since BC || AD)
Therefore,
∠ADC= ∠ABC = 100°, ∠ACD + ∠BCA = 80°, ∠CAD + ∠CAB = 80°.
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