Question

From the top of a hill 100m high the angle of depression of the top and the bottom of a tower are observed to be 30 and 45 respectively. Find the height of the tower.

Answer 1

$\tan(30 ) = (100 - x) \div 100$
$42.26$

Answer 2

Answer:

42.27 m

Step-by-step explanation:

Draw the following figure:

• Draw a straight line AB
• Take AB as the hill . AB is 100 m
• Draw another line parallel and shorter than AB beside AB
• Let the tower be DC
• Join D at a point on AB . Name the point E.

We have the following:

∠ACB = 45 ° .................(1)

∠ADE = 30 ° ...................(2)

Let DC be h

So AE = 100 - h

In triangle ACB,

tan 45° = AB/BC

==> AB / BC = 1

==> BC = AB

==> BC = 100 m

ED is parallel to BC

ED = BC = 100 m

ED = 100 m

In  triangle ACB,

tan 30° = AE / DE

==> AE / DE = 1 / √3

AE = 100 - h

==>( 100 - h) / 100 = 1 /√3

==> 100 - h = 100 / √3

==> h = 100 - 100 / √3

==> h = 100 - √3 *100 / 3

==> h = ( 300 - 1.732*100)/3

==> h = (300 - 173.2 )/3

==> h=126 .8 / 3

==> h = 42.2666 m

The height of the tower is 42 .67 m

Hope it helps you

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