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Geometry Level 4

In the above quadrilateral A B C D ABCD , A B = 3 m AB=3\text{ m} , B C = 4 m BC=4\text{ m} , C D = 12 m CD=12\text{ m} , A D = 13 m AD=13\text{ m} and angle A B C = 9 0 ∘ ABC=90^{\circ} . If the angle B A D BAD is expressible in the form tan ⁡ − 1 ( − a b ) \tan^{-1}(-\frac{a}{b}) , where a a and b b are coprime positive integers , find b − a b-a .


The answer is -23.

Arkajyoti Banerjee by Arkajyoti Banerjee , 20, India

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1 solution

Join A C AC . Then by Pythagorean Theorem, A C = 5 m AC=5m . Subsequently, tr. A C D ACD is right-angled at A C D ACD since it is a 5 m − 12 m − 13 m 5m-12m-13m triangle.

Let angle C A D = x CAD=x and angle B A C = y BAC=y . Then tan ⁡ x = 12 5 \tan x=\frac{12}{5} and tan ⁡ y = 4 3 \tan y=\frac{4}{3} . Now,

tan ⁡ ( x + y ) = tan ⁡ x + tan ⁡ y 1 − tan ⁡ x tan ⁡ y \tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}

⟹ tan ⁡ ( x + y ) = 12 5 + 4 3 1 − 12 5 × 4 3 \implies \tan(x+y)=\frac{\frac{12}{5} + \frac{4}{3}}{1-\frac{12}{5} \times \frac{4}{3}}

⟹ tan ⁡ ( x + y ) = − 56 33 ⟹ ( x + y ) = tan ⁡ − 1 ( − 56 33 ) \implies \tan(x+y)=-\frac{56}{33} \implies (x+y)=\tan^{-1}(-\frac{56}{33}) ⟹ a n g l e B A D = tan ⁡ − 1 ( − 56 33 ) \implies angleBAD = \tan^{-1}(-\frac{56}{33})

Clearly, a = 56 a=56 and b = 33 b=33 . So, b − a b-a gives us − 23 \boxed{-23}

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