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.

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.

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.

angle APB=cosarc-13/28

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.

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Have a look!Can you solve it?