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Upper primary school mathematics question UPQ64

In the figure below, not drawn to scale, ABC and AED are straight lines and BE = DE. Find ∠ BDC.

*Solution:*

∠ ADC = 180^{0} – 38^{0} – 72^{0} = 70^{0}

^{ }∠ BDC = ∠ EBD = 33^{0} (an isosceles triangle)

∠ BDC = 70^{0} – 33^{0} = 37^{0}

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