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 triangle ABC. (Solve this task without using Trigonometry)
Answers
Answer:
pls mark me as brianliets if you are satisfied
Step-by-step explanation:
The purpose of this task is to apply knowledge about triangles to calculate the sine and cosine of 30 and 60 degrees. Once the calculations of sin60∘ and cos60∘ have been made in part (b), there many ways to approach the calculations in part (d). Two methods are presented in the solution. Another nice approach is to reflect the picture from parts (a) and (b) over the line y=x. This interchanges the x and y coordinates. In terms of angles, the ray AC−→−, which makes a 60 degree angle with the x-axis, reflects to a ray making a 30 degree angle with the x-axis, giving us the expected relationship between sine and cosine of complementary angles.
Many variants are possible on the arguments given in the solution. For example, in part (c), reflection over the x-axis maps the circle to itself and interchanges the two 30 degree angles given in the picture. This means that reflection about the x-axis interchanges Q and R, making the x-axis the perpendicular bisector of QR¯¯¯¯¯¯¯¯. Another argument would show that △QPS and △RPS are congruent (via SAS for example). In part (a) also there are many different arguments for showing that CD←→ is perpendicular to AB←→.
Solution
Sides AB¯¯¯¯¯¯¯¯ and AC¯¯¯¯¯¯¯¯ both have unit length and this means that △ABC is isosceles with congruent angles ∠B and ∠C. Since m(∠A) is given as 60 this means that m(∠B)+m(∠C)=120 and so m(∠B)=m(∠C)=60. Hence △ABC is equilateral. Equilateral triangles have three lines of reflective symmetry, 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¯¯¯¯¯¯¯¯.
We apply the Pythagorean Theorem to the right triangle ADC, giving us
|AD|2+|CD|2=|AC|2.
We saw in part (a) that D is the midpoint of AB¯¯¯¯¯¯¯¯. Since B is on the unit circle, |AB|=1 and |AD|=12. We have |AC|=1 since C is on the unit circle. Plugging the values for |AD| and |AC| into the displayed formula gives |CD|=3√2. Since CD←→ is perpendicular to AB¯¯¯¯¯¯¯¯ we have
C=(12,3√2).
This means that sin60∘=3√2 and cos60∘=12.