Respuesta :

Answer:

[tex]\boxed{\boxed{\overline{AB}=16}}[/tex]

Step-by-step explanation:

In the triangle ABC, m∠ABC=90°. Hence, it is right angle triangle.

[tex]\tan \theta=\dfrac{p}{b}=\dfrac{AB}{64}[/tex]  --------1

In the triangle ABD, m∠ABD=90°. Hence, it is right angle triangle.

[tex]\cot (90-\theta)=\dfrac{b}{p}=\dfrac{4}{AB}[/tex]

As [tex]\cot (90-\theta)=\tan \theta[/tex], so

[tex]\tan \theta=\dfrac{b}{p}=\dfrac{4}{AB}[/tex]  ---------2

From equation 1 and 2,

[tex]\Rightarrow \dfrac{AB}{64}=\dfrac{4}{AB}[/tex]

[tex]\Rightarrow AB^2=64\times 4[/tex]

[tex]\Rightarrow AB^2=256[/tex]

[tex]\Rightarrow AB=\sqrt{256}[/tex]

[tex]\Rightarrow AB=16[/tex]


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