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April 20, 2014

April 20, 2014

Posted by **Sonny** on Monday, October 1, 2012 at 8:00pm.

- Math, Trig -
**Reiny**, Monday, October 1, 2012 at 8:46pmDraw a "side view"

Draw 3 points on the river, B , C, and D , B to the left of the points

draw a point A above B so that angle B is 90°

Join AC and AC , so that CD = 20

Mark angle ADC = 62.6° and angle ACB = 72.8°

label AB =h (h is the height) , BC = x

in triangle ABC:

tan 72.8° = h/x

h = xtan72.8°

in triangle ABD

tan62.6° = h/(x+20)

h = (x+20)tan62.6°

xtan72.8 = (x+20)tan62.6

xtan72.8 - xtan62.6) = 20tan62.6

x(tan72.8 - tan62.6) = 20tan62.6

x = 20tan62.6/(tan72.8 - tan62.6)

sub back into h = xtan72.8

h = 20(tan62.6)(tan72.8)/(tan72.8-tan62.6)

= ..... you do the button pushing

notice up to this point, I have not done any arithmetic or calculator work.

Also notice that the fact that the bridge was 1 km long did not enter the picture. It is important that you recognize what information is relevant and what is not.

BTW, I got 95.79 m

Alternate way:

in triangle ACD (not right-angled) , angle A = 10.2° , angle C = 107.2°

By the Sine Law:

AC/sin62.6 = 20/sin10.2

AC = 20sin62.5/sin10.2

in the right-angled triangle ABC

sin72.8 = h/AC

h = AC sin72.8 = (20sin62.6/sin10.2)(sin72.8) = 95.79

HOW ABOUT THAT ??

- Math, Trig -
**Sonny**, Monday, October 1, 2012 at 9:07pmThanks so much!

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