Find the value of sin 60° geometrically.

Open in App

Solution

Consider an equilateral ∆ABC with each side equal to 2a.
Then, each angle of ∆ABC = 60^o
From A, draw AD ⊥ BC.
Figure
Then, it is clear that BD = DC = a
Also, ∠ADB = 90^o
Now, in ∆ ADB, we have:
∠ADB + ∠DBA + ∠BAD = 180^o (angle sum property)
⇒ 90^o + 60^o + ∠ BAD = 180^o
⇒ ∠BAD = 180^o − 150^o = 30^o
∴ ∠BAD = 30^o

In ∆ADB, by Pythagoras theorem, we have:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow A D = \sqrt{A B^{2} - B D^{2}} = \sqrt{{(2 a)}^{2} - {(a)}^{2}} = \sqrt{4 a^{2} - a^{2}} = \sqrt{3} a$
∴ sin 60^o = $\frac{A D}{A B} = \frac{\sqrt{3} a}{2 a} = \frac{\sqrt{3}}{2}$

Suggest Corrections

Similar questions

(i) sin 60^{\circ} cos 30^{\circ} + cos 60^{\circ} sin 30^{\circ}

(i i) cos 30^{\circ} cos 60^{\circ} - sin 30^{\circ} sin 60^{\circ}

sin 60^{\circ} csc 60^{\circ}

Find the value of $s i n (60 ° + θ) - c o s (30 ° - θ) .$

Find the value of sin(60+ $θ$ ) - cos(30- $θ$ ).

Join BYJU'S Learning Program