Question

In the figure,find the value of DC

Answer 1

Answer:

Step-by-step explanation:

Given $AD=10\text{ m}$

Consider

DC = x m and BC = y m

In $\Delta ABC$ $\angle ACB=90^\circ\text{ and }\angle ABC=45^\circ$

$\therefore\frac{AC}{BC}=tan45^\circ\\\\\Rightarrow\frac{AC}{BC}=1\\\\\Rightarrow{AC}={BC}\\\\\Rightarrow10+x}=y$

$[tex]\Delta DBC$[/tex] $\angle BCD=90^\circ\text{ and }\angle DBC=\theta$

$\therefore\frac{BC}{DC}=cot\theta\\\\\Rightarrow\frac{10+x}{x}=cot\theta\\\\\Rightarrow\frac{10}{x}+1=cot\theta\\\\\Rightarrow\frac{10}{x}=cot\theta-1\\\\\Rightarrow{x}=\frac{10}{cot\theta-1}$

As per the figure the look like. If consider the theta=30°

then the value of

$x=\frac{10}{cot\theta-1}\\\\\Rihgtarrow{x}=\frac{10}{cot30^\circ-1}\\\\\Rihgtarrow{x}=\frac{10}{\sqrt{3}-1}\\\\\Rihgtarrow{x}=\frac{10}{1.73-1}\\\\\Rihgtarrow{x}=\frac{10}{.73}\\\\\Rihgtarrow{x}=13.69$

Answer 2

Answer:

this answer is wrong

answer is 60m