Question Number 218738 by Spillover last updated on 14/Apr/25
Answered by som(math1967) last updated on 15/Apr/25
Commented by som(math1967) last updated on 15/Apr/25
![area△ABC=(1/2)×a×b×sinA area △ABC=(1/2)×h×BC ∴(1/2)×ab×sinA=(1/2)×h×2RsinA [∵ BC=2RSinA] ∴ab=2Rh](https://www.tinkutara.com/question/Q218746.png)
$${area}\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×{b}×{sinA} \\ $$$${area}\:\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{h}×{BC} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}×{ab}×{sinA}=\frac{\mathrm{1}}{\mathrm{2}}×{h}×\mathrm{2}{RsinA} \\ $$$$\left[\because\:{BC}=\mathrm{2}{RSinA}\right] \\ $$$$\:\therefore{ab}=\mathrm{2}{Rh} \\ $$
Commented by Spillover last updated on 15/Apr/25
$${thanks} \\ $$
Answered by Spillover last updated on 15/Apr/25
