Given a square , and a point inside it, such that and .
what’s the answer/solution?
cos-1(-39/112)…. is it correct??
anurag jetley,,, u r right , even i get 110 degrees
apply cosine rule twice on angles PBA and PBC, where AB=BC. these 2 angles add up to 90 so sin of one angle is the cos of the other. use sqr sin + sqr cos = 1 to eliminate the unknown angle and get a biquadratic equation. Solve for the side AB. 3 sides of the triangle containing the able are known. compute the unknown angle using cosine law again applied to angle APB.
It can be either 69.6° or 110.4° by using coordinate geometry. I don’t know how to find the correct one out of these two.
I have calculated by using that resultant vector we use in parallelogram addition using cosine angles; I got the angle as approximately 69 degrees.
Reblogged this on IITJEE PREP and commented:
Have a look!Can you solve it?
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