Menu Close

If-a-flagstaff-subtends-equal-angles-at-4-points-A-B-C-and-D-on-the-horizontal-plane-through-the-foot-of-the-flagstaff-then-A-B-C-and-D-must-be-the-vertices-of-1-Square-2-Cyclic-quadrilateral




Question Number 15384 by Tinkutara last updated on 10/Jun/17
If a flagstaff subtends equal angles at 4  points A, B, C and D on the horizontal  plane through the foot of the flagstaff,  then A, B, C and D must be the  vertices of  (1) Square  (2) Cyclic quadrilateral  (3) Rectangle  (4) Parallelogram
$$\mathrm{If}\:\mathrm{a}\:\mathrm{flagstaff}\:\mathrm{subtends}\:\mathrm{equal}\:\mathrm{angles}\:\mathrm{at}\:\mathrm{4} \\ $$$$\mathrm{points}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{on}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{plane}\:\mathrm{through}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{flagstaff}, \\ $$$$\mathrm{then}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{must}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{vertices}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Square} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Cyclic}\:\mathrm{quadrilateral} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Rectangle} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Parallelogram} \\ $$
Answered by mrW1 last updated on 10/Jun/17
(2)
$$\left(\mathrm{2}\right) \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Sir please explain.
$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{explain}. \\ $$
Commented by mrW1 last updated on 10/Jun/17
h=height of flagstaff  d_A =distance of point A to the foot of flagstaff  θ_A =angle flagstaff subtends at point A  tan θ_A =(h/d_A )  tan θ_B =(h/d_B )  tan θ_C =(h/d_C )  tan θ_D =(h/d_D )  ∵θ_A =θ_B =θ_C =θ_D   ∴d_A =d_B =d_C =d_D =r  ⇒A, B,C,D have the same distance  to the foot of flagstaff, or they must  lie on a circle with center point at  the foot of flagstaff.  ⇒A, B,C,D are vertices of a cyclic  quadrilateral.
$$\mathrm{h}=\mathrm{height}\:\mathrm{of}\:\mathrm{flagstaff} \\ $$$$\mathrm{d}_{\mathrm{A}} =\mathrm{distance}\:\mathrm{of}\:\mathrm{point}\:\mathrm{A}\:\mathrm{to}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{flagstaff} \\ $$$$\theta_{\mathrm{A}} =\mathrm{angle}\:\mathrm{flagstaff}\:\mathrm{subtends}\:\mathrm{at}\:\mathrm{point}\:\mathrm{A} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{A}} =\frac{\mathrm{h}}{\mathrm{d}_{\mathrm{A}} } \\ $$$$\mathrm{tan}\:\theta_{\mathrm{B}} =\frac{\mathrm{h}}{\mathrm{d}_{\mathrm{B}} } \\ $$$$\mathrm{tan}\:\theta_{\mathrm{C}} =\frac{\mathrm{h}}{\mathrm{d}_{\mathrm{C}} } \\ $$$$\mathrm{tan}\:\theta_{\mathrm{D}} =\frac{\mathrm{h}}{\mathrm{d}_{\mathrm{D}} } \\ $$$$\because\theta_{\mathrm{A}} =\theta_{\mathrm{B}} =\theta_{\mathrm{C}} =\theta_{\mathrm{D}} \\ $$$$\therefore\mathrm{d}_{\mathrm{A}} =\mathrm{d}_{\mathrm{B}} =\mathrm{d}_{\mathrm{C}} =\mathrm{d}_{\mathrm{D}} =\mathrm{r} \\ $$$$\Rightarrow\mathrm{A},\:\mathrm{B},\mathrm{C},\mathrm{D}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{distance} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{flagstaff},\:\mathrm{or}\:\mathrm{they}\:\mathrm{must} \\ $$$$\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\mathrm{point}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{flagstaff}. \\ $$$$\Rightarrow\mathrm{A},\:\mathrm{B},\mathrm{C},\mathrm{D}\:\mathrm{are}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cyclic} \\ $$$$\mathrm{quadrilateral}. \\ $$
Commented by mrW1 last updated on 10/Jun/17
Commented by Tinkutara last updated on 10/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *