Find the value of tan60∘ geometrically.
Consider an equilateral ΔPQR. Since each angle in an equilateral triangle.
∴∠P=∠Q=∠R=600
Draw a perpendicular PS from P to QR.
Now, In ΔPQS and ΔPRS
∠PSQ=∠PSR[eachof900]
∠PQS=∠PRS[eachof600]
PS=PS[common]
∴ΔPQS≅ΔPRS(A.S.A.congruancy)
Let sides PQ=QR=RP=2x
Then, QS=SR=QR2=2x2=x
Now, In ΔPSQ
PQ2=QS2+SP2 By Pythagoras theorem
(2x)2=x2+PS2
PS2=3x2
PS=√3x
Now, In ΔPSQ
tanQ=PSQS
tan600=√3xx
tan600=√3
Hence, this is the answer.