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

The high level bridge, a railway bridge that crosses the Oldman River is over 1km long. From one point on the river, the angle of elevation of the top of the bridge is 62.6 degrees. From a point 20m closer to the bridge, the angle of elevation of the top of the bridge is 72.8 degrees. How high is the brige above the river, to the nearest meter? I am really confused by this question like I know what its asking but I don't know how to draw it out on paper so i can work it out.

- Math, Trig -
**Reiny**, Monday, October 1, 2012 at 8:46pm
Draw 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 9:07pm
Thanks so much!

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