In the given figure, if ∠ABC = 50o and ∠BDC = 40o, then ∠BCA is equal to [1]3 / 7a) 90oc) 40ob) 50od) 100
Question
In the given figure, if ∠ABC = 50o and ∠BDC = 40o, then ∠BCA is equal to [1]3 / 7a) 90oc) 40ob) 50od) 100
Solution
It seems like there's a mistake in your question as the figure is not provided. However, if we assume that ABC and BDC are linear pairs (which means they are adjacent and their outer sides form a straight line), then we can calculate ∠BCA.
Here are the steps:
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Since ∠ABC and ∠BDC are linear pairs, their sum should be 180o. So, we calculate ∠ABD = 180o - 50o - 40o = 90o.
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Now, ∠BCA is an exterior angle for triangle ABD, and the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.
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So, ∠BCA = ∠BAD + ∠ABD = 50o + 90o = 140o.
Please note that this solution is based on the assumption that ABC and BDC are linear pairs. If they are not, we would need the figure or more information to solve the problem.
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