Question

55. The length of the shadow of a tower
standing on level ground is found to be
2x m longer when the Sun's altitude is 30°
than when it was 45°. Prove that the height​

Answer 1

Step-by-step explanation:

Let h be the height of the tower.

In ΔABC,

BCAB=tan45∘

yh=1

h=y

In ΔABD,

BDAB=tan30∘

2x+yh=31

2x+y=3h

2x+h=3h

h=3−12x

h=(3+1)xmeters

Answer 2

$\huge \fbox \green{❤Answer:-}$

$\sf \colorbox{pink} {Let h be the height of the tower.}$

$In \: ΔABC,$

$⇒\frac{ AB }{BC} = tan \: 45°$

$⇒ \frac{h}{y} = 1$

$⇒h = y$

$In \: ΔABC,$

$⇒ \frac{AB}{BD} = tan \: 30°$

$⇒ \frac{h}{2x + y} = \frac{1}{ \sqrt{3} }$

$⇒2x + y = \sqrt{3h}$

$⇒2x + h = \sqrt{3h}$

$⇒h = \frac{2x}{ \sqrt{3} - 1 }$

$:h = ( \sqrt{3} + 1) \: x \: meters$

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$