Question

In a triangle ABC if angle C=90°,prove that / sin^2A+sin ^B=1

Answer 1

Answer:

sin2A=(BC/AB)2

and sin2B=(AC/AB)2

So, sin2A + Sin2B

=(BC/AB)2 + (AC/AB)2

=BC2 + AC2 / AB2 (Using pythagoras theorem we get BC2 + AC2 = AB2 )

=AB2/AB2

=1

Step-by-step explanation:

Answer 2

Trigonometry is a branch of mathematics that define the connection between measure of angles and side-lengths.

Explanation:

Given : In a triangle ABC if angle C=90°.

By angle sum property of triangle ,

∠ A+∠B+∠C = 180°

⇒ ∠ A+∠B+ 90°= 180°

⇒ ∠ A+∠B = 90°

⇒ ∠ A = 90° - ∠B   (1)

Since there is an identity in trigonometry : sin²x+cos²x=1

So for Angle A in triangle ABC , we have

sin²A+cos²A=1

⇒ sin²A+(cos A)²=1

⇒ sin²A+ (cos (90°-B))² =1   [From (1)]

⇒ sin²A+  (sin B)² =1    [∵ cos (90°-x)= sin x]

⇒ sin²A+ sin²B=1

Hence Proved.

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