For question no 25 (parallelogram) if we take angle ACD as Theta and angle DAC will be 150-theta, from a/sina = b/sinb we get sin theta = 1/4 , cos theta = sqrt15/4 we can solve it directly i.e 10 x 20 x sin(150 - theta)
bhai mughe 2025 wala exam dena hai then mughe kha se start krna chaiye kaise preparation konsa platform and kiski books and notes best honge plsss help bhai
For question no 25 (parallelogram) if we take angle ACD as Theta and angle DAC will be 150-theta,
from a/sina = b/sinb we get sin theta = 1/4 , cos theta = sqrt15/4
we can solve it directly i.e 10 x 20 x sin(150 - theta)
1st question where is it mentioned that the triangle to be right angled?
As BC as it’s diameter, it’s a semi circle, the angle inscribed in a semi circle is always 90 degrees
bhai mughe 2025 wala exam dena hai then mughe kha se start krna chaiye kaise preparation konsa platform and kiski books and notes best honge plsss help bhai
sincere thanks from my side sir!
sir please upload LR videos
Thank youuuuuuu
Ans B
Woah
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