If the non-base angle of angle of anisosceles triangle has a measure of 70o, what is themeasure of each base angle?

How many diagonals does a decagon have?

How many diagonals does a decagon have?

equissupnica7
2022-07-28
Answered

If the non-base angle of angle of anisosceles triangle has a measure of 70o, what is themeasure of each base angle?

How many diagonals does a decagon have?

How many diagonals does a decagon have?

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Marisa Colon

Answered 2022-07-29
Author has **18** answers

The sum of angles in any triangle is equal to 180 degrees. The two base angles are going to be equal in an isoscelestriangle.

Therefore 180-70 = 110

110 / 2 = 55 degrees

The number of diagonals in a polygon can be determined by theequation 1/2 * n * (n - 3) with n being the number of sides of saidpolygon. Therefore:

1/2 * 10 * (10 - 7) = 35

Therefore 180-70 = 110

110 / 2 = 55 degrees

The number of diagonals in a polygon can be determined by theequation 1/2 * n * (n - 3) with n being the number of sides of saidpolygon. Therefore:

1/2 * 10 * (10 - 7) = 35

asked 2022-06-20

Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vector of the point of concurrency in terms of position vectors of the vertices.

I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don't want to use vector equations for straight lines and then find the point of concurrency. That's like solving using coordinate geometry.

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ represent the sides $A,B,C$ respectively.

The angle bisectors are along $\frac{\overrightarrow{a}}{|\overrightarrow{a}|}}+{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}}+{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}}+{\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}$

Let the sides $AB,BC,CA$ be $x,y,z$. Let $AD$ be one of the angular bisector.

$\frac{BD}{CD}=\frac{x}{z}$

Hence

$D=\frac{x\overrightarrow{c}+z\overrightarrow{b}}{x+z}$

What should be the next step? Or is there a better method?

I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don't want to use vector equations for straight lines and then find the point of concurrency. That's like solving using coordinate geometry.

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ represent the sides $A,B,C$ respectively.

The angle bisectors are along $\frac{\overrightarrow{a}}{|\overrightarrow{a}|}}+{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}}+{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}}+{\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}$

Let the sides $AB,BC,CA$ be $x,y,z$. Let $AD$ be one of the angular bisector.

$\frac{BD}{CD}=\frac{x}{z}$

Hence

$D=\frac{x\overrightarrow{c}+z\overrightarrow{b}}{x+z}$

What should be the next step? Or is there a better method?

asked 2022-07-05

In a triangle $ABC$, take the tangent to the circumcircle of $ABC$ at $A$. Reflect this line through the angle bisector at $A$. prove that this reflected line is parallel to $BC$.

I'm looking for a quick and simple proof of this fact.

I'm looking for a quick and simple proof of this fact.

asked 2022-04-07

We have a $\mathrm{\u25b3}ABC$ and a $\mathrm{\u25b3}{A}_{1}{B}_{1}{C}_{1}$. The segments $CL$ and ${C}_{1}{L}_{1}$ are angle bisectors. If $\mathrm{\u25b3}ALC\sim \mathrm{\u25b3}{A}_{1}{L}_{1}{C}_{1}$, I should show that $\mathrm{\u25b3}ABC\sim \mathrm{\u25b3}{A}_{1}{B}_{1}{C}_{1}$.

From the similarity, we have $\frac{AL}{{A}_{1}{L}_{1}}}={\displaystyle \frac{CL}{{C}_{1}{L}_{1}}}={\displaystyle \frac{AC}{{A}_{1}{C}_{1}}$. The only way I see from here is to show that $\mathrm{\u25b3}LBC\sim \mathrm{\u25b3}{L}_{1}{B}_{1}{C}_{1}$. Is this necessary for the solution?

From the similarity, we have $\frac{AL}{{A}_{1}{L}_{1}}}={\displaystyle \frac{CL}{{C}_{1}{L}_{1}}}={\displaystyle \frac{AC}{{A}_{1}{C}_{1}}$. The only way I see from here is to show that $\mathrm{\u25b3}LBC\sim \mathrm{\u25b3}{L}_{1}{B}_{1}{C}_{1}$. Is this necessary for the solution?

asked 2022-05-10

I have the following problem: In a triangle ABC, the measure of the angle formed by the external angle bisectors of B and C is equal to twice the measure of the angle A. Find out the value of angle A.

Can anyone suggest how to approach this example?

Can anyone suggest how to approach this example?

asked 2022-06-25

Consider the hyperbolic 3-space $({\mathbb{H}}^{3},d{s}^{2})$ with

${\mathbb{H}}^{3}:=\{(x,y,z)\in {\mathbb{R}}^{3}|z>0\},\phantom{\rule{1em}{0ex}}d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}+d{z}^{2}}{{z}^{2}}$

Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$ .

${\mathbb{H}}^{3}:=\{(x,y,z)\in {\mathbb{R}}^{3}|z>0\},\phantom{\rule{1em}{0ex}}d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}+d{z}^{2}}{{z}^{2}}$

Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$ .

asked 2022-05-08

The lines which bisect an angle and the adjacent angle made by producing one of its arms are called the internal and external bisectors of an angle.

What does "external angle bisector" mean? How an angle can be bisected externally?

What does "external angle bisector" mean? How an angle can be bisected externally?

asked 2022-08-08

Find the locus of a point p whose distances from two fixed points A, A' are in a constant ratio 1:M, that is |PA| : |PA'| = 1:M. ( M >0)