trigo 6

• Aug 14th 2009, 05:51 AM
thereddevils
trigo 6
A rectangular lamina ABCD has sides AB=CD=2a and BC=AD=a . The midpoint of CD is M and the midpoint of BM is N . THe lamina is folded along BM such that the planes BMC and BMDA are perpendicular . For this folded lamina , find in terms of surds if necessary .

(a) AN
(b) AC
(c) angle ACB

This is what i did :

(a) BM = $a\sqrt{2}$, $BN =\frac{a\sqrt{2}}{2}$ , AB=2a

AM=AB=2a , cos angle MBC $= \frac{\sqrt{2}}{2} \Rightarrow$
angle MBC=45 degree , therefore ABN= 45 degree

$
AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
$

$
AN=\frac{\sqrt{10}}{2}a
$

(b) AMC=90 degree

MC=a , AM = 2a

$AC=\sqrt{(2a)^2+a^2}=a\sqrt{5}$

(c) ABC=90

tan angle ACB = 2a/a

ACB=arctan 2

= 63 degree 26 mins

CAn somepne pls check my working . Thanks a lot !!
• Aug 14th 2009, 10:29 AM
Amer
Quote:

Originally Posted by thereddevils
A rectangular lamina ABCD has sides AB=CD=2a and BC=AD=a . The midpoint of CD is M and the midpoint of BM is N . THe lamina is folded along BM such that the planes BMC and BMDA are perpendicular . For this folded lamina , find in terms of surds if necessary .

(a) AN
(b) AC
(c) angle ACB

This is what i did :

(a) BM = $a\sqrt{2}$, $BN =\frac{a\sqrt{2}}{2}$ , AB=2a

AM=AB=2a , cos angle MBC $= \frac{\sqrt{2}}{2} \Rightarrow$
angle MBC=45 degree , therefore ABN= 45 degree

$
AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
$

$
AN=\frac{\sqrt{10}}{2}a
$

(b) AMC=90 degree

MC=a , AM = 2a

$AC=\sqrt{(2a)^2+a^2}=a\sqrt{5}$

(c) ABC=90

tan angle ACB = 2a/a

ACB=arctan 2

= 63 degree 26 mins

CAn somepne pls check my working . Thanks a lot !!

$
AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
$

should be

$AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\cos 45^o$

$AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\left(\frac{1 }{\sqrt{2}}\right)$

$AN^2 = (2a)^2 + \left(\frac{a}{\sqrt{2}}\right)^2 - 2(2a)\left(\frac{a}{\sqrt{2}}\right)\left(\frac{1} {\sqrt{2}}\right)$

$AN^2 = 4a^2 + \left(\frac{a^2}{2}\right) - 2a^2$

$AN^2=2a^2+\frac{a^2}{2}$

$AN^2=\frac{5a^2}{2}\Rightarrow AN=a\sqrt{\frac{5}{2}}$

b) and c) is correct
• Aug 15th 2009, 06:16 PM
thereddevils
Quote:

Originally Posted by Amer
$
AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
$

should be

$AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\cos 45^o$

$AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\left(\frac{1 }{\sqrt{2}}\right)$

$AN^2 = (2a)^2 + \left(\frac{a}{\sqrt{2}}\right)^2 - 2(2a)\left(\frac{a}{\sqrt{2}}\right)\left(\frac{1} {\sqrt{2}}\right)$

$AN^2 = 4a^2 + \left(\frac{a^2}{2}\right) - 2a^2$

$AN^2=2a^2+\frac{a^2}{2}$

$AN^2=\frac{5a^2}{2}\Rightarrow AN=a\sqrt{\frac{5}{2}}$

b) and c) is correct

THANKS so i think the answer given for b and c in my book is wrong .