Asked by Rabin
Prove :Tan20+4sin20=square root 3
Answers
Answered by
Reiny
Let ABC be a right-angled triangle with ∠ABC = 60° and ∠CAB = 30°.
Let ABD be a right-angled triangle on the same side of AB
with ∠ABD = 40° and ∠DAB = 50°
Suppose that AC and BD intersect at E, and
that the length of BC is 1, so that the respective lengths of CA and AB are √3 and 2. Then
|AD| = |AB|sin 40° = 4 sin 20°cos 20° , (recall sin 2x = 2sinx cosx)
and
|AE| = |AD|sec 20° = |AB| cos 50°
sec 20° = 2 sin 40°
sec 20° = 4 sin 20°
However, |CE| = |BC|tan 20° = tan 20°
Therefore
tan 20° + 4 sin 20° = |CE| + |AE| = |AC| = √3
source:
http://www.math.toronto.edu/barbeau/olymon2010.pdf
page 9
Let ABD be a right-angled triangle on the same side of AB
with ∠ABD = 40° and ∠DAB = 50°
Suppose that AC and BD intersect at E, and
that the length of BC is 1, so that the respective lengths of CA and AB are √3 and 2. Then
|AD| = |AB|sin 40° = 4 sin 20°cos 20° , (recall sin 2x = 2sinx cosx)
and
|AE| = |AD|sec 20° = |AB| cos 50°
sec 20° = 2 sin 40°
sec 20° = 4 sin 20°
However, |CE| = |BC|tan 20° = tan 20°
Therefore
tan 20° + 4 sin 20° = |CE| + |AE| = |AC| = √3
source:
http://www.math.toronto.edu/barbeau/olymon2010.pdf
page 9
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