How can two angles be 90 degree in a right angled triangle ?

Question

If the ratio of sine of angles of a triangle is   1:1: sqrt(2)    then the ratio of square of its greatest
side to sum of the squares of other two sides is ...

 

I am not getting that if sin A=1 then A =90 as well as sin B =1 then B =90 , so how can it be possible , a bit confused in this ,plz clarify this ...

Answer 1

In the question, ratio of sine of angles is $1:1:\sqrt{2}$, not actual sine of angles.

Here just by looking, we can infer the three angles as 45, 45, 90 degrees (hence ratio of sine is $1:1:\sqrt{2}$). It is a right angled triangle, and so square of largest side is equal to sum of squares of other two sides, and so the ratio asked in question is 1.

In general, you can find angles as follows : Given ratio of sine as $1:1:\sqrt{2}$, let actual sine of angles be $k, k, \sqrt{2}k$.

Now in a triangle, A+B+C = 180 degrees i.e. A+B = 180-C

So $\sin (A+B) = \sin (180-c)$

$\implies \sin A\cos B + \cos A\sin B = \sin C$

$\implies k\sqrt{1-k^2} + \sqrt{1-k^2}*k = \sqrt{2}k$

$\implies 2\sqrt{1-k^2} = \sqrt{2}$

$\implies 4(1-k^2) = 2$

$\implies k = \frac{1}{\sqrt{2}}$

So sine of angles are $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1$, hence angles are 45, 45, 90 degrees.