Convert from radians to degrees:

$-\frac{3\pi}{8}$

$-\frac{3\pi}{8}$

babyagelesszj
2022-07-15
Answered

Convert from radians to degrees:

$-\frac{3\pi}{8}$

$-\frac{3\pi}{8}$

You can still ask an expert for help

Marisol Morton

Answered 2022-07-16
Author has **13** answers

$-\frac{3\pi}{8}$

$\pi \text{}rad={180}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{3}{8}\ast {180}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{3}{2\text{\u29f8}4}\ast \text{\u29f8}{90}^{\circ}{45}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{{135}^{\circ}}{2}$

$\pi \text{}rad={180}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{3}{8}\ast {180}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{3}{2\text{\u29f8}4}\ast \text{\u29f8}{90}^{\circ}{45}^{\circ}$

$-\frac{3}{8}\pi \text{}rad=-\frac{{135}^{\circ}}{2}$

Jeffrey Jordon

Answered 2022-07-26
Author has **2581** answers

asked 2022-06-14

Finding the slopes of internal or external angle bisectors, when given the slopes of the sides of a triangle, it's a simple question, the one you can easily find in any textbook of analytic geometry.

But the reverse problem : finding the slopes of the sides of a triangle, when only given the slopes of internal angle bisectors. Is it possible? Is there a straightforward formula for that?

But the reverse problem : finding the slopes of the sides of a triangle, when only given the slopes of internal angle bisectors. Is it possible? Is there a straightforward formula for that?

asked 2022-07-16

Internal angle bisector of $\mathrm{\angle}A$ of triangle $\mathrm{\Delta}ABC$, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of $\mathrm{\Delta}ABC$ then.

(a) $AD=\frac{2bc}{b+c}\mathrm{cos}\frac{A}{2}$

(b) $PQ=\frac{4bc}{b+c}\mathrm{sin}\frac{A}{2}$

(c) The triangle $\mathrm{\Delta}APQ$ is isosceles

(d) AP is HM of b and c

My approach is as follow

This is the rough image that I have drawn

$\frac{\mathrm{cos}\frac{A}{2}}{AD}=\frac{\mathrm{sin}{90}^{o}}{{b}_{1}}$

$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}$

$PD=AD\mathrm{tan}\frac{A}{2}$

$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}\Rightarrow \frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{AD\mathrm{tan}\frac{A}{2}}$

The official answer is a,b,c and d.

I am not able to approach from here

(a) $AD=\frac{2bc}{b+c}\mathrm{cos}\frac{A}{2}$

(b) $PQ=\frac{4bc}{b+c}\mathrm{sin}\frac{A}{2}$

(c) The triangle $\mathrm{\Delta}APQ$ is isosceles

(d) AP is HM of b and c

My approach is as follow

This is the rough image that I have drawn

$\frac{\mathrm{cos}\frac{A}{2}}{AD}=\frac{\mathrm{sin}{90}^{o}}{{b}_{1}}$

$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}$

$PD=AD\mathrm{tan}\frac{A}{2}$

$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}\Rightarrow \frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{AD\mathrm{tan}\frac{A}{2}}$

The official answer is a,b,c and d.

I am not able to approach from here

asked 2022-10-12

Let ${A}_{1},{B}_{1},{C}_{1}$ be the tangency points of the intersection of the excircles of a triangle $ABC$ with the sides $BC,CA,AB,$ respectively. Prove that the circumcircles of $AB{B}_{1}$ and $AC{C}_{1}$ meet on a bisector of $\mathrm{\angle}BAC.$

What I thought: Circle $(AB{B}_{1})$ has ${\omega}_{1}=b(s-a)$ and circle $(AC{C}_{1})$ has ${v}_{2}=c(s-a)$. Their radical axis is given by $-yc(s-a)+zb(s-a)$ which is equation of A angle bisector and we are done.

What I thought: Circle $(AB{B}_{1})$ has ${\omega}_{1}=b(s-a)$ and circle $(AC{C}_{1})$ has ${v}_{2}=c(s-a)$. Their radical axis is given by $-yc(s-a)+zb(s-a)$ which is equation of A angle bisector and we are done.

asked 2022-08-22

The exterior angle bisectors of $\mathrm{\angle}B$ and $\mathrm{\angle}C$ intersect on point $O$. $\mathrm{\angle}BOC=70\xb0$.

Find $\mathrm{\angle}OAC$.

asked 2022-07-04

The question is as follows:

Let $K(5,12)$, $L(14,0)$ and $M(0,0)$. The line $x+2y=14$ bisects angle $MLK$. Find equations for the bisectors of the angles $KML$ and $MKL$.

Any help will be truly appreciated!

Let $K(5,12)$, $L(14,0)$ and $M(0,0)$. The line $x+2y=14$ bisects angle $MLK$. Find equations for the bisectors of the angles $KML$ and $MKL$.

Any help will be truly appreciated!

asked 2022-07-07

Two lines: ${a}_{1}x+{b}_{1}y+{c}_{1}=0$ and ${a}_{2}x+{b}_{2}y+{c}_{2}=0$ are given. I know that the equation of its bisectors is $\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})}}=\pm \frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{({a}_{2}^{2}+{b}_{2}^{2})}}$ But I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming ${c}_{1},{c}_{2}$ both are of same sign, I know if ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0$ and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. $tan\theta =\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}$ and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming ${c}_{1},{c}_{2}$ both are of same sign IF ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0$ then if we take positive sign we get the obtuse angle bisector".

asked 2022-05-09

Suppose you have a pair of lines passing through origin, $a{x}^{2}+2hxy+b{y}^{2}=0$, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for 2 separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.