A cute geometry problem

Given a square ABCD, and a point P inside it, such that PA=4, PB=7 and PC=9.

Find \angle APB.


About shitikanth

I am a fourth year undergraduate student of Computer Science at IIT Kanpur. I love doing mathematics, and science. I also do a bit of coding in my free time.

8 responses to “A cute geometry problem”

  1. ankit says :

    what’s the answer/solution?

  2. anurag jetley says :

    cos-1(-39/112)…. is it correct??

  3. pun says :

    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.

  4. rajiv kumar says :

    angle APB=cosarc-13/28

  5. GDP says :

    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.

  6. Karthik says :

    I have calculated by using that resultant vector we use in parallelogram addition using cosine angles; I got the angle as approximately 69 degrees.

  7. iitjeeprepblog says :

    Reblogged this on IITJEE PREP and commented:
    Have a look!Can you solve it?

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