ABC is a triangle right angled at A. If the length of the hypotenuse is 2√2 times the length of the perpendicular from A on the hypotenuse ,then find angles B and C of the triangle.

Answer :

We form our diagram from given information , As  :



Given  : The length of the hypotenuse is 2 $\sqrt{2}$ times the length of the perpendicular from A on the hypotenuse .
So,
Let AD  =  x  , Then 

BC  =  2 $\sqrt{2}$x

We know $Sin \theta = \frac{Opposite}{Hypotenuse} , Cos \theta = \frac{Adjacent}{Hypotenuse}$ , So

$$\begin{gathered} \mathrm{In} ∆ \mathrm{ABC}\hspace{0.17em} \\ \mathrm{Cos} \mathrm{B} = \frac{\mathrm{AB}}{\mathrm{BC}} = \frac{\mathrm{AB}}{2\sqrt{2}\mathrm{x}} \\ \mathrm{Cos} \mathrm{B} = \frac{\mathrm{AB}}{2\sqrt{2}\mathrm{x}} --- ( 1 ) \\ \mathrm{In} ∆ \mathrm{ABD}\hspace{0.17em} \\ \mathrm{Sin} \mathrm{B} = \frac{\mathrm{AD}}{\mathrm{AB}} \\ \mathrm{Sin} \mathrm{B} = \frac{\mathrm{x}}{\mathrm{AB}} --- ( 2 ) \\ \mathrm{Multiply} \mathrm{equation} 1 \mathrm{and} 2 , \mathrm{and} \mathrm{get} \\ \mathrm{Sin} \mathrm{B} \mathrm{Cos} \mathrm{B} = \frac{\mathrm{AB}}{2\sqrt{2}\mathrm{x}} \times \frac{\mathrm{x}}{\mathrm{AB}} \\ \Rightarrow \mathrm{Sin} \mathrm{B} \mathrm{Cos} \mathrm{B} = \frac{1}{2\sqrt{2} } \\ \Rightarrow 2 \mathrm{Sin} \mathrm{B} \mathrm{Cos} \mathrm{B} = \frac{1}{\sqrt{2} } \\ \Rightarrow \mathrm{Sin} 2\mathrm{B} = \frac{1}{\sqrt{2} } \\ \Rightarrow \mathrm{Sin} 2\mathrm{B} = \mathrm{Sin} 45° \\ \Rightarrow 2\mathrm{B} = 45° \\ \Rightarrow \mathrm{B} = \frac{45° }{2} \\ \Rightarrow \mathrm{B} = 22.5° \\ \mathrm{We} \mathrm{know}, \\ \angle \mathrm{A} + \angle \mathrm{B} + \angle \mathrm{C} = 180° \\ \angle \mathrm{A} + 22.5°+ 90° = 180° \\ \angle \mathrm{A} = 67.5° \\ \mathrm{So}, \\ \mathbf{\angle }\mathbf{ }\mathbf{A}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{67}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{\angle }\mathbf{ }\mathbf{B}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{22}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$