Another way would be trigonometric, of course. If one draws a vertical line base to B, then we have a right △. The little △ to the left has base segment QA. Because of complimentary angles, △QAB is 90-10-80 so, [1.1] (𝒎 = AB) = 1.000000 … by my definition, to start; [2.1] (𝒒 = QA) = 𝒎 cos 80° [3.2] (𝒒 = QA) = 0.173648 [3.1] (𝒉 = QB) = 𝒎 sin 80° [2.2] (𝒉 = QB) = 0.984808 [4.1] (𝒔 = BC) = √((𝒎 + 𝒒)² + 𝒉²) … Pythagoras [4.1] (𝒔 = BC) = 1.532089 Then recognizing that the baseline of the overall △ is (𝒒 + 𝒔) [5.1] ∠D = arctan( 𝒉 / (𝒔 + 𝒒) ) ... where tan() = Opposite divided by Adjacent [5.2] ∠D = arctan( 0.984808 / (1.532089 ⊕ 0.173648) ) [5.3] ∠D = 30° Ta, da! Fun with trigonometry. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
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