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Question Number 113848 by deepraj123 last updated on 15/Sep/20

$$\mathrm{In}\mathrm{a}△ABC,\mathrm{\Sigma }{a}^2({\sin }^{2}B-{\sin }^{2}C)=$$

Answered by 1549442205PVT last updated on 16/Sep/20

$$\mathrm{We}\mathrm{have}\mathrm{\Sigma }{\mathrm{a}}^2({\sin }^2\mathrm{B}-{\sin }^2\mathrm{C})=$$$${\mathrm{a}}^2({\sin }^2\mathrm{B}-{\sin }^2\mathrm{C})+{\mathrm{b}}^2({\sin }^2\mathrm{C}-{\sin }^2\mathrm{A})$$$$+{\mathrm{c}}^2({\sin }^2\mathrm{A}-{\sin }^2\mathrm{B})$$$$\mathrm{From}\mathrm{sine}\mathrm{theorem}\mathrm{we}\mathrm{have}:$$$$\frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}=2\mathrm{R}\Rightarrow {\mathrm{a}}^2={4\mathrm{R}}^2{\sin }^2\mathrm{A}$$$${\mathrm{b}}^2={4\mathrm{R}}^2{\sin }^2\mathrm{B},{\mathrm{c}}^2={4\mathrm{R}}^2{\mathrm{sim}}^2\mathrm{C}.\mathrm{Replace}\mathrm{into}$$$$\mathrm{above}\mathrm{equality}\mathrm{we}\mathrm{get}\mathrm{\Sigma }{\mathrm{a}}^2({\sin }^2\mathrm{B}-{\sin }^2\mathrm{C})$$$$={4\mathrm{R}}^2[{\sin }^2\mathrm{A}({\sin }^2\mathrm{B}-{\sin }^2\mathrm{C})+{\sin }^2\mathrm{B}({\sin }^2\mathrm{C}-{\sin }^2\mathrm{A})$$$$+{\sin }^2\mathrm{C}({\sin }^2\mathrm{A}-{\sin }^2\mathrm{B})].$$$$(\mathrm{Put}\mathrm{x}={\sin }^2\mathrm{A},\mathrm{y}={\sin }^2\mathrm{B},\mathrm{z}={\sin }^2\mathrm{C})$$$$={4\mathrm{R}}^2[\mathrm{x}(\mathrm{y}-\mathrm{z})+\mathrm{y}(\mathrm{z}-\mathrm{x})+\mathrm{z}(\mathrm{x}-\mathrm{y})]$$$$={4\mathrm{R}}^2(\mathrm{xy}-\mathrm{xz}+\mathrm{yz}-\mathrm{yx}+\mathrm{zx}-\mathrm{zy})=0$$

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