Hi ya, the full question is the diagram shows two points B (the bottom of the triangle dead centre) and C the right hand side of the triangle at 45 degrees. The top of the triangle is T and i need to work ou the height between T & B, B is 20 metres from A which is the left hand side bottom of the triangle which is 37 degrees, i then need to calculate the distanced between B & C
hope this makes sense and thank you ever so much
Hi ya
Can you detail the calculation if the angles of the triangle were the other way round please the 37 degrees angle is on the right and 45 on the left all other info is the same
Right triangle ABT is isosceles because the two
acute angles of a right triangle are complementary
and if one is 45°, so is the other, angle BTC = 45°.
Since BT = AB, which is given to be 20 meters, so
the height BT is 20 meters (or as you spell it across
the pond, "metres").
Let x = BC
In right triangle BCT,
Put the tangent over 1:
Cross multiply:
x·tan(37°) = 20·1
x·tan(37°) = 20
x =
x = 26.54089643 meters
Edwin