If sin 60°= root3/2 then find ( a)cos 60°(b)tan60°(c) sec60°(d)cosec60°(e))osec60ot 60°
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Construct an equilateral triangle ABC , and find the midpoint of BC , calling it D . Since triangle ABC is equilateral, all of its angles are 60∘ , and line AD bisects angle A into two 30∘ angles. Hence triangle ABD is a right-angled triangle, with angle 60∘ at B. Since AB=BC and BC=2BD , we have AB=2BD . Now
AB2=AD2+BD2
(2BD)2=AD2+BD2
4BD2=AD2+BD2
3BD2=AD2
AD=(3–√)BD.
So sin60∘=ADAB=(3√)BD2BD=3√2.
(We can also show by this argument that cos60∘=12 , tan60∘=3–√ , sin30∘=12 , cos30∘=3√2 and tan30∘=13√
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