In Δ ABC, ∠B = 90° and AB : BC = 2 : 1. The value of sin A + cot C is

Given : $$\angle$$B = 90 and AB : BC = 2 : 1

To find : $$sin A + cot C$$ = ?

Solution : Let AB = $$2x$$ and BC = $$x$$

In right $$\triangle$$ABC

=> AC = $$\sqrt{(AB)^2 + (BC)^2}$$

=> AC = $$\sqrt{4x^2 + x^2} = \sqrt{5x^2}$$

=> AC = $$\sqrt{5}x$$

=> $$sin A = \frac{BC}{AC}$$

=> $$sin A = \frac{x}{\sqrt{5}x} = \frac{1}{\sqrt{5}}$$

=> $$cot C = \frac{BC}{AB}$$

=> $$cot C = \frac{x}{2x} = \frac{1}{2}$$

Now, $$sin A + cot C$$

= $$\frac{1}{\sqrt{5}} + \frac{1}{2}$$

= $$\frac{2+\sqrt{5}}{2\sqrt{5}}$$