# A cute geometry problem

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

Find $\angle APB$.

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.

### 6 responses to “A cute geometry problem”

1. ankit says :

2. anurag jetley says :

cos-1(-39/112)…. is it correct??

• swaraj says :

anurag jetley,,, u r right , even i get 110 degrees

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.