1@ Race - still I am unable to find the θ. how did u get it to be 20°
let ABC be a triangle with P a point inside it such that <PAB=Ï€/18
<PBA=Ï€/9 <PCA=Ï€/6, <PAC=2Ï€/9
Q.<PBC=?
(a)Ï€/9
(b)2Ï€/9
(c)Ï€/3
(d)none
Q.Triangle ABC is___________
a)equilateral
b)isosceles
c)rt. angled
d)none
Q.Which is true
a)BC>AC
b)BC<AB
c)AC>AB
d)BC=AC
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UP 0 DOWN 0 0 4
4 Answers
1
from there it follows;
i got θ=20o and the answers are Q1-C; Q2-B; Q3-C
11sin(80-θ)/sinθ = 2.sin20.sin40/sin10 = cos60-cos20 / sin10
= cos10-2cos20.cos10 / 2.sin10.cos10
= cos10-(cos30+cos10) / sin20
=> sin(80-θ)/sinθ= cos30/sin20 = sin60/sin20 => θ = 20. (here -ve sign has to be neglected since from the figure it is clear that θ is acute angle)
hope it's clear now :)