ABC is an right angle triangle and PRQ is also right angle triangle an...
Proof:
Given: ABC is a right angle triangle with angle C as 90 degrees.
PRQ is also a right angle triangle with angle R as 90 degrees.
Angle B and angle Q are acute angles.
sin B = sin Q.
To prove: angle B = angle Q.
Proof by contradiction:
Assume that angle B is not equal to angle Q.
Case 1: angle B > angle Q
Let's assume angle B is greater than angle Q.
Since ABC is a right angle triangle, the sum of all angles in a triangle is 180 degrees. Therefore, angle B + angle C + angle A = 180 degrees.
In triangle ABC, angle C = 90 degrees. Substituting the value of angle C, we get angle B + 90 degrees + angle A = 180 degrees.
Simplifying the equation, we have angle B + angle A = 90 degrees.
In triangle PRQ, angle R = 90 degrees. Therefore, angle Q + 90 degrees + angle P = 180 degrees.
Simplifying the equation, we have angle Q + angle P = 90 degrees.
Given that sin B = sin Q, we can write the equation as sin B = sin (90 - angle P).
Using the identity sin (90 - x) = cos x, we have sin B = cos angle P.
Since angle B > angle Q, sin B > sin Q. But from our assumption, sin B = sin Q.
This is a contradiction, as sin B cannot be greater than sin Q when angle B > angle Q.
Therefore, our assumption that angle B > angle Q is false.
Case 2: angle B < angle="" />
Let's assume angle B is less than angle Q.
Using similar logic as in Case 1, we can arrive at a contradiction.
Therefore, our assumption that angle B < angle="" q="" is="" />
Conclusion:
Since both assumptions, angle B > angle Q and angle B < angle="" q,="" lead="" to="" contradictions,="" we="" can="" conclude="" that="" angle="" b="" must="" be="" equal="" to="" angle="" />
Hence, angle B = angle Q.
This proves that if sin B = sin Q, then angle B = angle Q.
ABC is an right angle triangle and PRQ is also right angle triangle an...
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