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Question-9677




Question Number 9677 by ridwan balatif last updated on 23/Dec/16
Commented by geovane10math last updated on 23/Dec/16
An acute angle can not measure 129°  51° + 78° = 129°
$${An}\:{acute}\:{angle}\:{can}\:{not}\:{measure}\:\mathrm{129}° \\ $$$$\mathrm{51}°\:+\:\mathrm{78}°\:=\:\mathrm{129}° \\ $$
Commented by ridwan balatif last updated on 24/Dec/16
ooh i′m so sorry, i′m forget to say that point   D is in the middle of line BC
$$\mathrm{ooh}\:\mathrm{i}'\mathrm{m}\:\mathrm{so}\:\mathrm{sorry},\:\mathrm{i}'\mathrm{m}\:\mathrm{forget}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{point}\: \\ $$$$\mathrm{D}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\mathrm{line}\:\mathrm{BC} \\ $$
Answered by mrW last updated on 23/Dec/16
α=78°  β=51°  a=AC=12 cm  L=AD  ((L×sin β)/(a×sin (α+β)))=(1/2)  L=((sin (α+β)×a)/(2sin β))=((sin (129°)×12)/(2sin 51°))=6 cm
$$\alpha=\mathrm{78}° \\ $$$$\beta=\mathrm{51}° \\ $$$$\mathrm{a}=\mathrm{AC}=\mathrm{12}\:\mathrm{cm} \\ $$$$\mathrm{L}=\mathrm{AD} \\ $$$$\frac{\mathrm{L}×\mathrm{sin}\:\beta}{\mathrm{a}×\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{L}=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)×\mathrm{a}}{\mathrm{2sin}\:\beta}=\frac{\mathrm{sin}\:\left(\mathrm{129}°\right)×\mathrm{12}}{\mathrm{2sin}\:\mathrm{51}°}=\mathrm{6}\:\mathrm{cm} \\ $$

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