Albert1
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How Do You Calculate the BC Angle in a Triangle?
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Opalg
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[sp]
By Pythagoras, $\overline{DE} = 6$. With angles $\alpha,\;\beta$ as marked in the diagram, $\tan\alpha = \frac34$. Therefore $$\tan\beta = \tan(45^\circ - \alpha) = \frac{1-\frac34}{1 + \frac34} = \frac17.$$ So $\overline{BD} = \frac87$, and (Pythagoras) $\overline{AB} = \dfrac{40\sqrt2}7.$ It follows that $\overline{AC} = 40\sqrt2$, and (Pythagoras again) $\overline{BC} = \dfrac{400}7.$[/sp]
By Pythagoras, $\overline{DE} = 6$. With angles $\alpha,\;\beta$ as marked in the diagram, $\tan\alpha = \frac34$. Therefore $$\tan\beta = \tan(45^\circ - \alpha) = \frac{1-\frac34}{1 + \frac34} = \frac17.$$ So $\overline{BD} = \frac87$, and (Pythagoras) $\overline{AB} = \dfrac{40\sqrt2}7.$ It follows that $\overline{AC} = 40\sqrt2$, and (Pythagoras again) $\overline{BC} = \dfrac{400}7.$[/sp]
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Albert1
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thanks for your participation, your answer is correct !
this time using geometry only ,hope someone can do it
this time using geometry only ,hope someone can do it
Albert1
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