In ∆ABC angleABC is 30°.AD is the median on BC.AngleADC is 45°.Find angleBAC.
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angle ABC is 30 and angle add is 45 and angle ADB is 135 by linear pair and angle BAD is 15 angle back is 30 because AD is median
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∠ABC=30°
∠ADC=45°
inΔBAD
∠ADC=∠ABC+∠BAD[∵exterior ∠ of a Δ is equal to the sum of its opposite interior∠s]
∠BAD=45°-30°
∠BAD=15°
∠ADC=45°
inΔBAD
∠ADC=∠ABC+∠BAD[∵exterior ∠ of a Δ is equal to the sum of its opposite interior∠s]
∠BAD=45°-30°
∠BAD=15°
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