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Question
Observe the given construction, if \(\angle{BAE} = 60^{\circ}\), then \(\angle{CAF}\) will be
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Solution
\(\angle{BAE} = 60^{\circ}\) [Given]
AF is angle bisector of \(60^{\circ}\)
\(\Rightarrow \angle{BAF} = 30^{\circ}\)
and AC is angle bisector of \(30^{\circ}\)
\(\angle{BAC}=\angle {CAF}\)
\(\Rightarrow \angle{BAC} =\angle {CAF} = 15^{\circ}\)
\(\angle{BAE} = 60^{\circ}\) [Given]
AF is angle bisector of \(60^{\circ}\)
\(\Rightarrow \angle{BAF} = 30^{\circ}\)
and AC is angle bisector of \(30^{\circ}\)
\(\angle{BAC}=\angle {CAF}\)
\(\Rightarrow \angle{BAC} =\angle {CAF} = 15^{\circ}\)
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