Question-47831

Question Number 47831 by somil last updated on 15/Nov/18

Commented by somil last updated on 16/Nov/18

i k n o w

Answered by $@ty@m last updated on 15/Nov/18

\tan 45 = \frac{P Q}{Q R}

\Rightarrow P Q = Q R = x, s a y \dots (1)

\tan 30 = \frac{x}{x + 30}

\frac{1}{\sqrt{3}} = \frac{x}{x + 30}

x \sqrt{3} = x + 30

x = \frac{30}{\sqrt{3} - 1}

x = 15 (\sqrt{3} + 1) = 15 \times 2.732

x = 40.98 c m

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18

t a n 45^{o} = \frac{P Q}{Q R} = 1 s o P Q = Q R

t a n 30^{o} = \frac{P Q}{Q D} = \frac{P Q}{Q R + R D} = \frac{P Q}{P Q + R D} = \frac{1}{\sqrt{3}}

P Q \sqrt{3} = P Q + R D

P Q \sqrt{3} = P Q + 30

P Q \sqrt{3} - P Q = 30

P Q (\sqrt{3} - 1) = 30

P Q = \frac{30}{\sqrt{3} - 1} = \frac{30 (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \frac{30 (\sqrt{3} + 1)}{3 - 1} = 15 (\sqrt{3} + 1)

P \overset{}{Q} = Q R = 15 (\sqrt{3} + 1) = 40.98 c m