Question

1. the triangle ABC is right-angled with right angle at C. the height from the corner C divides the hypotenuse into two parts with one length 1 and the other 4. What is the circumference of the triangle.
2. given a right triangle with catheter lengths a and b, determine and enter the length of the bisector at the right angle.
3. a rectangle has a diagonal length d. the acute angle between the diagonals is x. given that cos (x) = p, determine and state the length of the shorter sides of the rectangle as a function of p
4. Given a > 1, what is the biggest solution to: (1/x-1) ≥ a/(x+1)
5. Of cos x = p, and pi/2<x<pi. Give a expression for tan x

Answer 1

1)

$a^2=h^2+1^2$

$b^2=h^2+4^2$

$5^2=a^2+b^2$

$5^2=1+b^2+16+b^2$

$2h^2=25-17=8$

$h^2=4$

$h=2$

so $a^2=2^2+1=5$

$a=2.236067977$

$b^2=2^2+16=20$

$b=4.472135955$

circumference

$=2.236067977+4.472135955+5=11.70820393$

2)

$BC=a, CA=b, AB=\sqrt{a^2+b^2}$

CD is the angle bisector

let's draw square CEDF and CE=h

so,

$BE=a-h , CD=h\sqrt{2}$

Now, ABC and DBE are similar triangles.

$\frac{b}{a}=\frac{h}{a-h}$

$h=\frac{ab}{a+b}$

length of angle bisector is given by

$CD=h\sqrt{2}$

$CD=\frac{ab\sqrt{2}}{a+b}$

3)

shortest sides of the rectangle are AB and DC

using trigonometric ratio in triangle ABC.

$<B=90\degree$

now,

$cosx=\frac{adjacent}{hypotenuse}$

$cosx=\frac{AB}{AC}$

$cosx=\frac{AB}{d}$

now, $p=\frac{AB}{d}$

$AB=pd$

length of shortest side $AB=p \times d$

4)

$a>1$

$(\frac{1}{x}≥\frac{a}{x+1}$

$\frac{1}{x}-x+\frac{1}{x}-1≥a$

$\frac{x}{x}-x+\frac{1}{x}-1≥a$

$1-x+\frac{1}{x}-1≥a$

$-x+\frac{1}{x}≥a$

$\frac{1}{x}-x≥a$

$\frac{1-x^2}{x}≥a$

5)

$cos x=p$

$\frac{1}{secx}=p$

$sec x=\frac{1}{p}$

$sec^2 x=(\frac{1}{p})^2$

By identity

$sec^2x-tan^2x=1$

$tan^2x=sec^2x-1$

$=\frac{1}{p^2}-1$

$=\frac{1-p^2}{p^2}$

$tan^2x=\frac{1-p^2}{p^2}$

$tanx =\sqrt{\frac{1-p^2}{p^2}}$

$tanx=\frac{\sqrt{1-p^2}}{p}$