Two lines AB and CD intersect each other at a point O such that ∠AOC:∠AOD=5:7. Find all the angles.

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Solution

Let ∠AOC=5k and ∠AOD=7k, where k is a constant.
Here ∠AOC and ∠AOD form a linear pair.
∠AOC+∠AOD=180°
5k+7k=180°
12k=180°
k=15°
∠AOC=5k=5×15=75°
∠AOD=7k=7×15=105°
Now, ∠BOD=∠AOC=75° [Vertically opposite angles]
∠BOC=∠AOD=105° [Vertically opposite angles]

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Two lines AB and CD intersect each other at a point O such that ∠AOC:∠AOD=5:7. Find all the angles.

Let ∠AOC=5k and ∠AOD=7k, where k is a constant.Here ∠AOC and ∠AOD form a linear pair.∠AOC+∠AOD=180°5k+7k=180°12k=180°k=15°∠AOC=5k=5×15=75°∠AOD=7k=7×15=105°Now, ∠BOD=∠AOC=75° [Vertically opposite angles]∠BOC=∠AOD=105° [Vertically opposite angles]

Let ∠AOC=5k and ∠AOD=7k, where k is a constant.
Here ∠AOC and ∠AOD form a linear pair.
∠AOC+∠AOD=180°
5k+7k=180°
12k=180°
k=15°
∠AOC=5k=5×15=75°
∠AOD=7k=7×15=105°
Now, ∠BOD=∠AOC=75° [Vertically opposite angles]
∠BOC=∠AOD=105° [Vertically opposite angles]