Hi! I have a trig problem. What we know: AB = 12 mm CD = 6 mm DA = 20 mm ABC = 110° BCD = 130° What I would like to know: DAB = ? ° Thanks a lot in advance for any help! Regards
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angle DAB $\displaystyle \approx 50.48{}^{\circ}$
Thanks but did you determinate it algebraically? How?
Law of sines in triangle ABE and again in triangle DCE $\displaystyle \frac{\sin E}{6}=\frac{\sin 50^{\circ }}{d}$ $\displaystyle \frac{\sin E}{12}=\frac{\sin 110^{\circ }}{d+20}$ where d = ED