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October 24, 2014

October 24, 2014

Posted by **garry** on Thursday, January 15, 2009 at 9:55pm.

- trig -
**bobpursley**, Thursday, January 15, 2009 at 10:17pmlet h be height. Let d be the distance from the first observation to the bridge.

h/d=tan62.6

and

h/(d-20)=tan72.8

in the first equation

d=h ctn62.6

second equation

d=h ctn72.8 +20tan72.8

set these two equations equal, and solve for h.

- trig -
**Reiny**, Thursday, January 15, 2009 at 10:38pmor....

Consider the triangle formed with 20 m as the base and the two lines forming the angles of elevation to the top of the bridge.

62.6 will be the interior angle and 72.8 would be an exterior angle at the 20 m base.

That would make the interior angle at the base 107.2 and the small angle at the top where the lines meet at the bridge equal to 10.2 degrees.

by the Sine Law we can find the length of the line formed by the 72.8 degree line of observation.

this then becomes the hypotenuse of a right angled triangle where h is the height of the bridge, the side found above is the hypotenuse and the base angle is 72.8.

A simple sine ratio will find the height.

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