Category: Coordinate Geometry

Question Number 171512 by infinityaction last updated on 16/Jun/22 Answered by som(math1967) last updated on 17/Jun/22 $$\stackrel{⌢}{DL}=\stackrel{⌢}{LM}=\stackrel{⌢}{MB}$$$$\therefore \mathrm{\angle }DAL=\mathrm{\angle }LAM=\mathrm{\angle }MAB=30$$$${ar}.\bigtriangleup{ALM}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}{sin}\mathrm{30}×\mathrm{8}=\mathrm{16}{cm}^{\mathrm{2}} \…

Question Number 39955 by rahul 19 last updated on 13/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18 Answered…

Question Number 39860 by ajfour last updated on 12/Jul/18 Commented by ajfour last updated on 12/Jul/18 $$OA=BP,OT=1,\mathrm{\&}$$$$\mathrm{\angle }OTB=\mathrm{\angle }BTP\left(givenconditions\right).$$ Answered by ajfour last…

Question Number 105273 by ajfour last updated on 27/Jul/20 Commented by ajfour last updated on 27/Jul/20 $$GivenP(h,k)andB(p,q)$$$$TofindimageI(r,t)ofB(p,q)$$$$inthemirrorlineAT.$$ Answered by…

Question Number 170323 by cortano1 last updated on 20/May/22 Answered by mr W last updated on 20/May/22 $$\frac{AC}{\sin 18°}=\frac{AD}{\sin 12°}\dots \left(i\right)$$$$\frac{\sin (\alpha +18°)}{DB}=\frac{\sin \alpha }{AD}\dots \left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \…