Question

A 8 In the adjoining figure, AB = 4 m and ED = 3 m. If sin a. 12 and cos B - find the length of BD. 13 Hint. sin a 3 . tan a 3 5 E 4 m 3 m B с Also cos B pa 12 13 5 4 tần B 12​

Answer 1

Step-by-step explanation:

Length of BD = 12.53 m.

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Answer 2

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Given:

Given:Here,

AB = 4 m and ED = 3 m.

$sin\alpha = \frac{3}{5} , cos B= \frac{12}{13}$

$cos B=\frac{12}{13}\Rightarrow tan B=\frac{5}{12}cosB=1312⇒tanB=125$

To find:

Length of BD = ?

Solution:

$sin\alpha = \frac{3}{5}$

$\frac{AB}{AC} =\frac{3}{5}$

$\frac{4}{AC} =\frac{3}{5}$

$AC =\frac{5\times 4}{3}$

$AC= \frac{20}{3}$

$AB^2+BC^2=AC^2AB2+BC2=AC2$

$BC^2=AC^2-AB^2BC2=AC2−AB2$

     

$= (\frac{20}{3})^2 +(4)^2=(320)2+(4)2$

     

$=\frac{400}{9}+16=9400+16$

       

$=\frac{400-144}{9}=9400−144$

$BC^2=\frac{256}{9}BC2=9256</p><p>BC=\sqrt{\frac{256}{9} }BC=9256</p><p>BC=\frac{16}{3}BC=316</p><p>Cos B = \frac{12}{13} \Rightarrow tab B =\frac{5}{12}CosB=1312⇒tabB=125</p><p>\frac{ED}{CD} =\frac{5}{12}CDED=125</p><p>\frac{3}{CD} =\frac{5}{12}CD3=125</p><p>CD = \frac{3\times 12}{5}CD=53×12</p><p>CD =\frac{36}{5}CD=536</p><p>Therefore, BD = BC + CD</p><p>                        = \frac{16}{3} +\frac{36}{5}316+536</p><p>                        = \frac{16\times 5+3\times 36}{5\times 3}5×316×5+3×36</p><p>                        = \frac{188}{15}15188</p><p>                        = 12.53 m$

Length of BD = 12.53 m