Question

in triangle abc right angled at B angle ACB is equal to theta a b is equal to 2 centimetre and BC is equal to 1 centimetre find the values of sin squared theta + tan squared theta​

Answer 1

If AB= 2 cm & BC = 1 cm & ∠ACB = θ in right-angled triangle ABC, then the value of sin squared theta + tan squared theta is $\frac{24}{5}$.

Step-by-step explanation:

We are given ∆ABC is right-angled triangle at B i.e.,

∠B = 90°

Also given, ∠ACB = θ, AB = 2 cm & BC = 1 cm

Applying Pythagoras theorem in right-angled ∆ABC, we get

AC = $\sqrt{AB^2 + BC^2}$ = $\sqrt{2^2 + 1^2}$ = √5 cm

Now, using the trigonometry property of triangle in ∆ABC, we have  

sin θ = $\frac{Perpendicular}{Hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{2}{\sqrt{5}}$…… (i)

and,

tan θ = $\frac{Perpendicular}{Base}$ = $\frac{AB}{BC}$ = $\frac{2}{1}$ ……. (ii)

Thus, substituting from (i) & (ii), we get

The value of sin²θ + tan²θ as,

= [$\frac{2}{\sqrt{5}}$]² + [2]²

= $\frac{4}{5}$ + 2

= $\frac{4 + 20}{5}$

= $\frac{24}{5}$

Hope this is helpful!!!!!

Answer 2

Step-by-step explanation:

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