The perpendicular bisector 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

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Supoilign1964 · Accepted Answer

Step 1  The Sides of AB and AC have unit length  Hence, △ABC is isosceles with Congruent angles ∠B and ∠C NSNK Since m(∠A) is given as 60 this means that,  m(∠B)+m(∠C)=120  and so, m(∠B)+m(∠C)=60  Therefore △ABC is equilateral  Equilateral triangles have three lines of reflective symmetry in which the lines joining each vertex to the midpoint of the opposite side.  This means that CD is a line of symmetry for △ABC and so CD is perpendicular to AB.  Step 2  By applying Pythagorean Theorem to the right triangle △ADC  |AD|2+|CD|2=|AC|2  We Know that  D is the midpoint of AB.  B is on the unit circle,  Hence |AB|=1 and |AD|=12  Since C is unit circle, We have |AC|=1  Plugging the values for |AD| and |AC| into the formula gives,  |CD|=32. Since CD is perpendicular to AB .  Therefore, C(12, 32)  sin⁡60∘=32 and cos⁡60∘=12

The perpendicular bisector \overline{AB} in the right triangle \triangle ABC

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