Question

In ∆ABC and ∆PQR,if angleA and angleP are acute angles such that sinA=sinP then prove that angleA=angleP

Answer 1

Answer:

First of all we can understand that trignometric ratios are only possible for the right triangles.

So, let us suppose that angle B and angle Q are right angles of their respective triangles.

Gurther answer is in the pic. Thank you

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\huge {\sf {GIVEN\:THAT\: :-}}$

• ∠ A and ∠ P are acute angles
• sin A = sin P

$\huge {\sf {PROVE\:THAT\: :-}}$

• ∠ A = ∠ P

$\huge {\sf {SOLUTION\: :-}}$

sin A = sin P

We know that,

$\huge{\boxed {\tt Sin \theta = \dfrac {Perpendicular}{Hypotenuse}}}$

$\hookrightarrow \tt {\dfrac {AB}{AC}=\dfrac {PQ}{PR}}$

Therefore,

$\hookrightarrow \tt {\dfrac {AB}{PQ}=\dfrac {AC}{PR}}$

So, the sides of both the triangles are proportional.

Therefore, By SSS,

$\tt {\triangle ABC \sim \triangle PQR}$

Therefore,

$\huge{\boxed {\tt∠ A = ∠ P}}$

Because of corresponding parts of similar triangles.

___________