Question

Everyone is dying to answer my question.. xD!!

Here's one more..

Answer 1

HERE IS YOUR ANSWER

IN TRIANGLE ABD

TAN60 ° = AB / BD

root 3 = h/x

h=x root 3. ..........eq1

CD = EB

IN TRIANGLE AEC

tan45° = AE /x

AE = x

50 -h = x

h= 50+ x .........eq2

ON EQUATING eq 1 and 2

x root3 = x+50

x root 3 - x = 50

×(root 3 -1) =50

x = 50/root 3-1 × root 3 +1 /root3 +1

x = 25 (root 3 +1 )

h = x + 50

h = 50 +25 (root +1 )

HOPE IT HELPS YOU!

Answer 2

$\huge{Answer}$

h=50+25(√3+1)

Explanation

In ∆ABD

$tan60 = \frac{ab}{bd}$

$\sqrt{3} = \frac{h}{x}$

$h = x \sqrt{3} - - - - - (1)$

CD=EB

In ∆AEC

$tan45 = \frac{ae}{x}$

AE=X

50-h=x

h=50+x-----------(2)

From 1 and 2

$x \sqrt{3} = x + 50$

$x \sqrt{3} - x = 50$

$x( \sqrt{3} - 1) = 50$

$x = \frac{50}{ \sqrt{3} - 1 \times \frac{ \sqrt{3} + 1}{ \sqrt{3 + 1} } }$

$x = 25( \sqrt{3 } + 1)$

$h = 50 + 25( \sqrt{3} + 1)$