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Question-23138




Question Number 23138 by math solver last updated on 26/Oct/17
Commented by math solver last updated on 26/Oct/17
q.4?
$$\mathrm{q}.\mathrm{4}? \\ $$
Answered by ajfour last updated on 26/Oct/17
h_C =bsin A≥ c  h_B =csin A ≥b  so    bcsin^2 A ≥ bc  ⇒  sin A=1  and b=c  ⇒ ∠A=90° , ∠B=∠C=45° .
$${h}_{{C}} ={b}\mathrm{sin}\:{A}\geqslant\:{c} \\ $$$${h}_{{B}} ={c}\mathrm{sin}\:{A}\:\geqslant{b} \\ $$$${so}\:\:\:\:{bc}\mathrm{sin}\:^{\mathrm{2}} {A}\:\geqslant\:{bc} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:{A}=\mathrm{1}\:\:{and}\:{b}={c} \\ $$$$\Rightarrow\:\angle{A}=\mathrm{90}°\:,\:\angle{B}=\angle{C}=\mathrm{45}°\:. \\ $$
Commented by ajfour last updated on 26/Oct/17
h_c  is the perpedicular from C  on AB and is = bsin A  as    sin A=(h_c /b) and likewise..  it is given in question h_c  ≥ c  so   bsin A ≥ c  similarly  csin A ≥ b .
$${h}_{{c}} \:{is}\:{the}\:{perpedicular}\:{from}\:{C} \\ $$$${on}\:{AB}\:{and}\:{is}\:=\:{b}\mathrm{sin}\:{A} \\ $$$${as}\:\:\:\:\mathrm{sin}\:{A}=\frac{{h}_{{c}} }{{b}}\:{and}\:{likewise}.. \\ $$$${it}\:{is}\:{given}\:{in}\:{question}\:{h}_{{c}} \:\geqslant\:{c} \\ $$$${so}\:\:\:{b}\mathrm{sin}\:{A}\:\geqslant\:{c} \\ $$$${similarly}\:\:{c}\mathrm{sin}\:{A}\:\geqslant\:{b}\:. \\ $$
Commented by math solver last updated on 26/Oct/17
Commented by math solver last updated on 26/Oct/17
same sol. is given in book but   i could not understand the 1st 2   lines eq. i.e h _c  and h_b  ?
$$\mathrm{same}\:\mathrm{sol}.\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{book}\:\mathrm{but}\: \\ $$$$\mathrm{i}\:\mathrm{could}\:\mathrm{not}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{1st}\:\mathrm{2}\: \\ $$$$\mathrm{lines}\:\mathrm{eq}.\:\mathrm{i}.\mathrm{e}\:\mathrm{h}\:_{\mathrm{c}} \:\mathrm{and}\:\mathrm{h}_{\mathrm{b}} \:? \\ $$
Commented by math solver last updated on 26/Oct/17
thank you sir !
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$

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