How Do You Calculate the BC Angle in a Triangle?

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Discussion Overview

The discussion revolves around calculating the angle BC in a triangle, specifically using trigonometric and geometric methods. Participants explore different approaches to arrive at the solution.

Discussion Character

  • Technical explanation, Homework-related

Main Points Raised

  • One participant uses trigonometric identities and Pythagorean theorem to derive lengths and angles in the triangle, concluding with the length of side BC.
  • Another participant acknowledges the correctness of the first post and expresses a desire for a solution using only geometric methods, indicating a preference for an alternative approach.

Areas of Agreement / Disagreement

There is agreement on the correctness of the initial calculations, but a disagreement remains regarding the method of solution, as one participant seeks a purely geometric approach.

Contextual Notes

The discussion does not resolve the method preference, and the geometric approach remains unaddressed.

Who May Find This Useful

Readers interested in triangle geometry, trigonometry, and problem-solving methods in mathematics may find this discussion relevant.

Albert1
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Mathematics news on Phys.org
[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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thanks for your participation, your answer is correct !
this time using geometry only ,hope someone can do it
 

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