Question

In a right angled isosceles triangle ABC,sinA+sinB+sinC=​

Answer 1

Answer:

Perimeter of Triangle/ Hypotenuse.

Step-by-step explanation:

sin =opposite side /hypotenuse

sin A +sin B +sin C

BC/AC+AC/AC+AB/AC

(BC+AC+AB)/AC

Perimeter of triangle/Hypotenuse.

Hope this bring a smile in your face

Answer 2

In a right angled isosceles triangle ABC , sinA + sinB + sinC = 1 + √2

Given :

ABC is a right angled isosceles triangle

To find :

The value of sinA + sinB + sinC

Solution :

Step 1 of 2 :

Find the angles

Since ABC is a right angled isosceles triangle

So three angles are 90°, 45° , 45°

Without any loss of generality we assume that

A = 90°, B = 45° , C = 45°

Step 2 of 2 :

Find the value of sinA + sinB + sinC

sinA + sinB + sinC

= sin 90° + sin 45° + sin 45°

$\displaystyle \sf{ = 1 + \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } }$

$\displaystyle \sf{ = 1 + \frac{1 + 1}{ \sqrt{2} } }$

$\displaystyle \sf{ = 1 + \frac{2}{ \sqrt{2} } }$

$\displaystyle \sf{ = 1 + \sqrt{2} }$

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