3 years ago
The perpendicular bisector of hypotenuse AB in the right triangle △ ABC forms the triangle with the area 3 times smaller than the area of △ ABC. Find the measures of acute angles in △ ABC.(Solve this t without using Trigonometry)
1 Answers
Best Answer
bjhb Staff answered 3 years ago
bjhb Staff answered 3 years ago
Let the length be CB =l and AC=h in △ ABC.
So, by Pythagoras theorem,
AB = sqrt(h^(2)+l^(2))
Also, P is a perpendicular bisector of AB.
So, AP=PB
Hence, AP = PB =(sqrt(h^(2)+l^(2)))/(2)
Step2
Here, we see that △ ABC is ~ilar to △ APQ.
Since, both are right angle triangles.
So, ratio of their sides is
(△ ACB)/(△ APQ)=(3)/(1)
l=(l)/(sqrt(3))h
From the ratio of side one can clearly say that ∠ A = 30°, ∠ B = 60°.
So, by Pythagoras theorem,
AB = sqrt(h^(2)+l^(2))
Also, P is a perpendicular bisector of AB.
So, AP=PB
Hence, AP = PB =(sqrt(h^(2)+l^(2)))/(2)
Step2
Here, we see that △ ABC is ~ilar to △ APQ.
Since, both are right angle triangles.
So, ratio of their sides is
(△ ACB)/(△ APQ)=(3)/(1)
l=(l)/(sqrt(3))h
From the ratio of side one can clearly say that ∠ A = 30°, ∠ B = 60°.