Question

In ∆ABC, A = 20, B = 80, C = 80. D lies on AC such that angle DBC = 70. E lies on AB such that ECB = 50. Using Trigonometric Version of Ceva's Theorem deduce that angle BDE = 10. All measures are in degree.

Answer 1

Answer on Question #48567, Math, Trigonometry

In $\Delta ABC$, $A = 20$, $B = 80$, $C = 80$. D lies on AC such that angle DBC = 70. E lies on AB such that $ECB = 50$. Using Trigonometric Version of Ceva's Theorem deduce that angle BDE = 10. All measures are in degree.

Solution.

By applying Trigonometric Version of Ceva's Theorem for $\Delta DEC$ we get next equality:

$$\frac {\sin \angle E D B}{\sin \angle B D C} \cdot \frac {\sin \angle D C B}{\sin \angle B C E} \cdot \frac {\sin \angle C E B}{\sin \angle B E D} = 1.$$

From the task we know that $\angle CEB = \angle BCE = 50$, and $\angle BDC = 30$, $\angle DCB = 80$. The angle $\angle BED = 170 - \angle EDB$, which could be found from $\Delta BED$. Then we substitute this angles in the equality and obtain the new one:

$$\frac {\sin \angle E D B}{\sin 3 0} \cdot \frac {\sin 8 0}{\sin (1 7 0 - \angle E D B)} = 1.$$

Hence, $2\sin 80 \cdot \sin \angle EDB = \sin (170 - \angle EDB)$, transforming this equality we obtain:

$$\cos (8 0 - \angle E D B) - \cos (8 0 + \angle E D B) = \sin (1 7 0 - \angle E D B) \Rightarrow$$ $$\sin (1 0 + \angle E D B) - \sin (1 0 - \angle E D B) = \sin (1 0 + \angle E D B).$$

Finally, $\sin (10 - \angle EDB) = 0$ hence $\angle EDB = 10$.

Answer: EDB=10.

www.assignmentexpert.com