Question-185770

Question Number 185770 by Mingma last updated on 27/Jan/23

Answered by mahdipoor last updated on 27/Jan/23

Δ A D B : \frac{A B}{s i n (180 - (20 + x))} = \frac{B D}{s i n (x)}

Δ B D C : \frac{B D}{s i n 30} = \frac{B C}{s i n (180 - (50 + 30))}

Δ A B C : \frac{A B}{s i n 70} = \frac{B C}{s i n (180 - (70 + 70))}

\Rightarrow\Rightarrow B D = \frac{s i n (x)}{s i n (20 + x)} A B = \frac{s i n (30)}{s i n (80)} B C =

\frac{s i n (30)}{s i n (80)} \times \frac{s i n (40)}{s i n (70)} A B

\Rightarrow \frac{s i n (x)}{s i n (x + 20)} = \frac{s i n (40) . s i n (30)}{s i n (70) . s i n (80)} \Rightarrow

c t g (x) = \frac{s i n (70) s i n (80)}{s i n (40) s i n (30) s i n (20)} - c t g (20)

\Rightarrow x = 10