Geometry -- Trigonometry
Geometry
Level
pending
In triangle PAT. ∠P=36º, ∠A=56º, and PA=10. Point U and G lie on sides TP and TA respectively so that PU=AG=1. Let M and N be the midpoints of segments PA and UG respectively. What is the degree measure of the acute angle formed by lines MN and PA?
The answer is 80.
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1 solution
Create a rectangular coordinate system with P as the origin.
We get:
P (0,0)
W (5,0)
A (10, 0)
U (cos∠TPA, sin∠TPA)
=> U (cos 36º, sin 36º)
=> G (10-cos∠56º, sin∠56º)
Therefore, N ((10 + cos36º -cos56º)/2, (sin36º + sin 56º)/2)
=> tan x = (sin36º + sin 56º)/(cos 36º + cos 56 º)
(In order to simplify the above equation, we need to use Sum to Product Formula)
sin x + sin y = 2sin ((x+y)/2) cos ((x-y)/2)
cos x + cos y = 2cos ((x+y)/2) cos ((x-y)/2)
=>
tan x = (2sin (36º) cos (10º))/2/(10 + 2cos (36º) cos (10º))/2 -5
tan x = sin36º + sin56º/ cos36º + cos 56º
tan x = 2 sin46º cos10º/2 cos 46º sin 10º
tan x = sin46º/cos46º
tan x = cot 10º = tan 80º
The acute angle is 80º