Question

In the triangle below, what is the tangent of 60*?

Answer 1

Answer:

Step-by-step explanation:

We have 3 triangles here, all of them right triangles. The first one is the biggest one, triangle ABC. We know the side length AB = 14, and we know the hypotenuse length, BC = 30. We can use that, along with Pythagorean's Theorem, to solve for side length AC:

$30^2=14^2+(AC)^2$ and

$900-196=(AC)^2$ and

$704=(AC)^2$ so

$AC=8\sqrt{11}$

We also have the smallest right triangle, triangle ABD, which has the base of x, our unknown. Side BD is our unknown, length x; side AD is the height, length y (also unknown); and side AB is the hypotenuse, length 14. We will come up with a system of equations for these 2 right triangles, using Pythagorean's Theorem, and then solve for x using substitution.

For the medium-sized triangle. triangle ADC, the hypotenuse is side AC with a measure of 8√11, the height is side AD length y, and the base is side DC length 30 - x. Putting those values into Pythagorean's Theorem:

$(30-x)^2+y^2=(8\sqrt{11})^2$ and

$900-60x+x^2+y^2=704$ and

$x^2-60x+y^2=-196$***  That's starred because it's important and we will come back to it.

The smallest triangle in Pythagorean's Theorem:

$14^2=x^2+y^2$ and solving that for y-squared:

$y^2=196-x^2$. We will now sub that value for y-squared into the starred equation for y-squared and solve for x:

$x^2-60x+196-x^2=-196$ and

-60x = -392 so

x = 6.5