Question

If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.

Answer 1

SOLUTION :  

Given : ∠A and ∠P are acute angles tan A = tan P

To Prove :  ∠A =∠P

Let  right angled triangle be ∆ACP, ∠C = 90°

tan θ = perpendicular /base

With reference to ∠P,  

perpendicular  = AC  

base = CP  

Hypotenuse = AP

With reference to ∠A,  

perpendicular  = CP

base = AC

Hypotenuse = AP

tan A = P/B  = PC / AC

tan P = P/B  = AC/PC

∴ tan A = tan P    (Given)

PC / AC = AC/PC

PC = AC  

∠A = ∠P

[In triangle angle opposite to equal sides are equal]

Hence, ∠A = ∠P

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Answer 2

Given : ∠A and ∠P are acute angles tan A = tan P

To Prove :  ∠A =∠P

Let  right angled triangle be ∆ACP, ∠C = 90°

tan θ = perpendicular /base

With reference to ∠P,  

perpendicular  = AC  

base = CP  

Hypotenuse = AP

With reference to ∠A,  

perpendicular  = CP

base = AC

Hypotenuse = AP

tan A = P/B  = PC / AC

tan P = P/B  = AC/PC

∴ tan A = tan P    (Given)

PC / AC = AC/PC

PC = AC  

∠A = ∠P

[In triangle angle opposite to equal sides are equal]

Hence, ∠A = ∠P