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Question Number 211392 by CrispyXYZ last updated on 08/Sep/24
△ABC. cos C=((sin A + cos A)/2)=((sin B + cos B)/2).  Find cos C.
$$\bigtriangleup{ABC}.\:\mathrm{cos}\:{C}=\frac{\mathrm{sin}\:{A}\:+\:\mathrm{cos}\:{A}}{\mathrm{2}}=\frac{\mathrm{sin}\:{B}\:+\:\mathrm{cos}\:{B}}{\mathrm{2}}. \\ $$$$\mathrm{Find}\:\mathrm{cos}\:{C}. \\ $$
Answered by Frix last updated on 08/Sep/24
Rectangular triangle  cos A =sin B ∧ sin A = cos B ⇒ C=90°
$$\mathrm{Rectangular}\:\mathrm{triangle} \\ $$$$\mathrm{cos}\:{A}\:=\mathrm{sin}\:{B}\:\wedge\:\mathrm{sin}\:{A}\:=\:\mathrm{cos}\:{B}\:\Rightarrow\:{C}=\mathrm{90}° \\ $$

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