Question

8 In the adjoining figure, AB = 4 m and ED = 3 m.
If sin a = 3/5; and cos B = 12/13, find the length of BD.
Hint. sin a = 3/5= tan a =3/4
Also cos B = 12/13 = tan B = 5/12


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Answer 1

Length of BD = 12.53 m

Step-by-step explanation:

Given:

Here,

AB = 4 m and ED = 3 m.

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

$cos B=\frac{12}{13}\Rightarrow tan B=\frac{5}{12}$

To find:

Length of BD = ?

Solution:

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

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

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

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

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

$AB^2+BC^2=AC^2$

$BC^2=AC^2-AB^2$

       $= (\frac{20}{3})^2 +(4)^2$

       $=\frac{400}{9}+16$

       $=\frac{400-144}{9}$

$BC^2=\frac{256}{9}$

$BC=\sqrt{\frac{256}{9} }$

$BC=\frac{16}{3}$

$Cos B = \frac{12}{13} \Rightarrow tab B =\frac{5}{12}$

$\frac{ED}{CD} =\frac{5}{12}$

$\frac{3}{CD} =\frac{5}{12}$

$CD = \frac{3\times 12}{5}$

$CD =\frac{36}{5}$

Therefore, BD = BC + CD

                        = $\frac{16}{3} +\frac{36}{5}$

                        = $\frac{16\times 5+3\times 36}{5\times 3}$

                        = $\frac{188}{15}$

                        = 12.53 m

Length of BD = 12.53 m