Can you find the value of theta? |

[@holyshit922]

Drop perpendicular from P to BC
Let Q be intersection of ths perpendicular and BC
Let R be the point on PQ such that PR = 2PQ
Triangle RCP is isosceles
then if we are able to show that triangle RPB is equilateral the problem is solved
But it seems that it leads to nowhere
Yes this sine rule solution is quite easy and i probably will get it

[@oxbmaths]

Nice problem and surprisingly neat answer!
My contribution (using sₙ=sin(n°), cₙ=cos(n°), tₙ=tan(n°)):
s₂₀s₃₀s₈₀ / (s₁₀s₄₀+ s₂₀s₃₀c₈₀)
= s₂₀s₃₀s₈₀ / ((s₄₀+ s₂₀s₃₀)c₈₀) [ ∵ s₁₀=c₈₀]
= s₂₀s₃₀s₈₀ / ((s₄₀+ (s₈₀- s₄₀)/2)c₈₀)
= s₂₀s₃₀s₈₀ / ((s₈₀+ s₄₀)/2)c₈₀)
= s₂₀s₃₀s₈₀ / (2s₆₀c₂₀/2)c₈₀)
= s₂₀s₃₀s₈₀ / (c₂₀c₃₀c₈₀) [∵ s₆₀=c₃₀]
= t₂₀t₃₀t₈₀ (× t₄₀/t₄₀)
= t₄₀t₂₀t₈₀ · t₃₀ · t₅₀ [∵ t₅₀=1/t₄₀]
= t₆₀ · t₃₀ · t₅₀ [∵ tan(60°-x) tan(x) tan(60°+x) = tan(3x)]
= t₅₀ [∵ t₆₀=1/t₃₀]

[@Why553-k5b_1]

theta=180-70-30-a°
a°=180-10-b°
b°=360-120-(150-theta)
b°=240-(150-theta)
b°=90+theta
a°=170-(90+theta)
a°=80-theta
theta=80-a=80-(80-theta)=theta
W H A T T H E F U C K

[@rakesh.a.c]

Because theta-p = -90 is same as p-theta = 90. You are solving using the same equation twice. You cannot form 2 distinct equations using theta and p for the above problem. If you do you get the thing you got above. It's not wrong but we are not solving the problem either. It's like saying 2=2 or x=x. We are not getting anywhere with those equations.

[@robertjarman3703]

There are not that many possible solutions. Especially given that the numbers are in multiples of ten. Brute forcing it would probably be faster.