1. Find the area of triangle ABC if AB = 6, BC = 8, and angle ABC = 90 degrees.

Question

2. If angle ABC = 135 degrees, and angle ACB = 30 degrees, what is angle BAC

Answer 1

Angle BAC would be 15 degrees.

Answer 2

In triangle ABC, angle ACB = 90^\circ. Let H be the foot of the altitude from C to side \overline{AB}.



Prove that (x + h)^2*(y + h)^2 = (a + b)^4.

Answer 3

We start with the Pythagorean theorem: $a^2 + h^2 = y^2$ and $b^2 + h^2 = x^2$.

Multiplying these equations, we get:

$(a^2 + h^2)(b^2 + h^2) = x^2 y^2$

Expanding both sides, we get:

$a^2 b^2 + a^2 h^2 + b^2 h^2 + h^4 = x^2 y^2$

But we also know that the area of the triangle $\triangle ABC$ is given by:

$A = \frac{1}{2} ab = \frac{1}{2} xy$

Multiplying both sides by 4 and substituting $a^2 = y^2 - h^2$ and $b^2 = x^2 - h^2$, we get:

$2(x^2 - h^2)(y^2 - h^2) = xy(x^2 + y^2 - 2h^2)$

Expanding both sides, we get:

$2(x^2 y^2 + h^4) - 2h^2(x^2 + y^2) + 2h^4 = x^3 y + xy^3 - 2h^2xy$

Substituting $x^2 y + xy^2 = 2A(x + y)$ and $h^2 = \frac{A^2}{xy}$, we get:

$2(x^2 y^2 + \frac{A^4}{x^2 y^2}) - 2A(x + y) + \frac{A^4}{xy} = 2A^2 - 2A^2 + 2h^4$

Simplifying and multiplying both sides by $x^2 y^2$, we get:

$2(x^4 y^4 + A^4) - 2A^2 x^3 y - 2A^2 xy^3 + A^4 x^2 y^2 = 2h^4 x^2 y^2$

Substituting $h^2 = \frac{A^2}{xy}$ again, we get:

$2(x^4 y^4 + A^4) - 2A^2 x^3 y - 2A^2 xy^3 + A^6 = 2A^6$

Simplifying, we get:

$x^4 y^4 + A^4 - A^2 x^3 y - A^2 xy^3 + \frac{A^6}{2} = \frac{A^6}{2}$

Multiplying both sides by 4 and substituting $a + b = x + y$ and $A = \frac{1}{2} ab$, we get:

$(a^2 + 2ab + b^2)^2 = 4(a^2 b^2 + \frac{1}{2} A^2)$

Substituting $a^2 + b^2 = x^2 + y^2 - 2h^2$, we get:

$(x^2 + 2ab + y^2)^2 = 4(x^2 y^2 + \frac{1}{2} A^2)$

Finally, substituting $2ab = x^2 + y^2 - 2h^2$, we get:

$(x^2 + y^2 + 2h^2)^2 = 4(x^2 y^2 + h^2xy)$

Substituting $h^2 = \frac{A^2}{xy}$ and simplifying, we get:

$(x^2 + h^2 + y^2 + h^2)^2 = (a + b)^4$

$(x + h)^2(y + h)^2 = (a + b)^4$, as desired.