Make a triangle ABC with angle A at 45 degrees Make the height AD of this triangle where D falls on the BC side BD=6 and CD=4 Find AD

To find the height AD of the triangle, we can use the trigonometric ratio for the sine of angle A.

In triangle ABC, BD = 6 and CD = 4. Let's denote AD as h.

Since angle A is 45 degrees, angle B is 90 degrees - 45 degrees = 45 degrees as well. Therefore, triangle ABC is an isosceles right triangle.

Using the Pythagorean theorem, we can find the length of side BC: BC^2 = BD^2 + CD^2 BC^2 = 6^2 + 4^2 BC^2 = 36 + 16 BC^2 = 52 BC = √52 BC ≈ 7.211

Since triangle ABC is an isosceles right triangle, BD = CD = BC/√2 BD = CD = 7.211/√2 BD = CD ≈ 5.099

Now, let's consider triangle ABD. We know that angle A is 45 degrees, angle D is 90 degrees, and angle B is 45 degrees. Using the sine ratio: sin(A) = opposite/hypotenuse sin(45) = h/BD 1/√2 = h/5.099

Solving for h: h = (1/√2) * 5.099 h = 5.099/√2 h = 5.099 * √2/2 h ≈ 3.605

Therefore, AD ≈ 3.605