is isosceles with . Given that and , find correct to four decimal places.
Triangle
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Since ∠ B E C = 1 8 0 − 4 0 − 3 0 − 4 0 = 7 0 , △ E B C is isosceles with B C = B E .
∠ C D B = 1 8 0 − 3 0 − 3 0 − 4 0 = 8 0 ∘
Let B C = B E = 1
By sine law on △ E B C , we have
sin 4 0 C E = sin 7 0 B C
C E = sin 7 0 sin 4 0
By sine law on △ D B C , we have
sin 7 0 D C = sin 8 0 B C
D C = sin 8 0 sin 7 0
sin 3 0 B D = sin 8 0 B C
B D = sin 8 0 sin 3 0
By cosine law on △ E D C , we have
D E 2 = ( sin 7 0 sin 4 0 ) 2 + ( sin 8 0 sin 7 0 ) 2 − 2 ( sin 7 0 sin 4 0 ) ( sin 8 0 sin 7 0 ) ( cos 4 0 )
D E ≈ 0 . 6 1 5 1 3 2 1 4 8
By sine law on △ D E B , we have
sin 8 0 sin 3 0 sin ∠ D E B = 0 . 6 1 5 1 3 2 1 4 8 sin 3 0
∠ D E B ≈ 2 4 . 3 7 3 7