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Given-ABC-where-BC-a-AC-b-and-AB-c-If-A-60-Find-the-value-of-1-b-c-a-c-1-c-b-a-b-




Question Number 195517 by cortano12 last updated on 04/Aug/23
   Given ΔABC where BC= a,      AC = b and AB = c . If ∠ A= 60°     Find the value of     (1+(b/c)+(a/c))(1+(c/b)+(a/b)).
$$\:\:\:\mathrm{Given}\:\Delta\mathrm{ABC}\:\mathrm{where}\:\mathrm{BC}=\:\mathrm{a},\: \\ $$$$\:\:\:\mathrm{AC}\:=\:\mathrm{b}\:\mathrm{and}\:\mathrm{AB}\:=\:\mathrm{c}\:.\:\mathrm{If}\:\angle\:\mathrm{A}=\:\mathrm{60}° \\ $$$$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\left(\mathrm{1}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{a}}{\mathrm{c}}\right)\left(\mathrm{1}+\frac{\mathrm{c}}{\mathrm{b}}+\frac{\mathrm{a}}{\mathrm{b}}\right). \\ $$$$ \\ $$
Answered by horsebrand11 last updated on 05/Aug/23
 ((b+a)/c) = ((cos (((A−B)/2)))/(sin ((C/2))))   ((a+c)/b)= ((cos (((C−A)/2)))/(sin ((B/2))))
$$\:\frac{\mathrm{b}+\mathrm{a}}{\mathrm{c}}\:=\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{\mathrm{C}}{\mathrm{2}}\right)} \\ $$$$\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}=\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{C}−\mathrm{A}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{\mathrm{B}}{\mathrm{2}}\right)} \\ $$$$\: \\ $$

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