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Geometry: Coordinates and Angles

In the triangle with vertices G( –5, –8), H(3, 0), and K(9, 0)
the angle at H is ° to 3 dps.

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Worked Solution

We can use the following formula to find angle H:
Cosine Rule: h2 = k2 + g2 – 2 kg cos(H)  or  cos(H) =
k2 + g2h2
---
2 kg

First we will need the lengths of the sides g, h and k.
As we are given the coordinates of GHK we can use Pythagoras' Theorem to find these lengths
but we also need their squares in the Cosine Rule:
g2 = (9 – 3)2 + (0 – 0)2 = 36 so g = √36 = 6
h2 = (9 – (–5))2 + (0 – (–8))2 = 260
k2 = (3 – (–5))2 + (0 – (–8))2 = 128 so k = √128 = 11.31371
Substituting these in the formula for cos(H) we have:
128 + 36 – 260
---
2 √128 √36
= –0.70711
Angle H = arccos( –0.70711) = 135°


© MEI Produced by Dr Ron Knott, revised 28 January 2005
tom . button [AT] mei . org . uk
Test reference: GeomCoordsAngs.html?ref=4081