it is a problem of geometry n fundamental results tell me d solution............

## in triangle ABC, A is a right angle and D is a point on AC such that BD bisects B angle. if angle BDC = 100 degree , calculate angle C

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# 3 Answers

The sum of the three angles in any triangle is 180 degrees.

Triangle ABC is a right triangle with the right angle at vertex A. Point D is on the line AC so that the like BD bisects angle B

So line BD divides the triangle ABC into two triangles, ABD and DBC, and the angle at vertex B is the same for both triangles. Let's give that angle a name, X.

We also know that the angle at vertex D in triangle DBC is 100 degrees. This makes the angle at vertex D in triangle ABD 80 degrees because the two angles must add to 180 degrees.

At this point we know two of the angles in triangle ABD add up to 170 degrees, so the third angle, which we called X, is equal to 10 degrees.

Now we know two of the angles in triangle DBC, one is X (= 10 degrees) and the other is 100 degrees. So the angle at vertex C must be 70 degrees.

Triangle ABC is a right triangle with the right angle at vertex A. Point D is on the line AC so that the like BD bisects angle B

So line BD divides the triangle ABC into two triangles, ABD and DBC, and the angle at vertex B is the same for both triangles. Let's give that angle a name, X.

We also know that the angle at vertex D in triangle DBC is 100 degrees. This makes the angle at vertex D in triangle ABD 80 degrees because the two angles must add to 180 degrees.

At this point we know two of the angles in triangle ABD add up to 170 degrees, so the third angle, which we called X, is equal to 10 degrees.

Now we know two of the angles in triangle DBC, one is X (= 10 degrees) and the other is 100 degrees. So the angle at vertex C must be 70 degrees.

Hi Rodela;

Total degrees of a triangle...180

ABC total 180

A=90

B=B

C=180-(B+90)=90-B

180=(90)+(B)+(90-B)

BDC total 180

D=100

B=(1/2)B

C=90-B

180=(100)+[(1/2)B]+(90-B)

Let's subtract one equation from the other...

180=(90)+(B)+(90-B)

-(180=(100)+[(1/2)B]+(90-B))

0=-10+[(1/2)B]

Let's add 10 to both sides...

10+0=10-10+[(1/2)B]

10=(1/2)B

Let's multiply both sides by 2...

(2)(10)=(2)(1/2)B

20=B

A=90

B=20

C=180-(90+20)=180-110=70

The quickest way to do these is usually to draw them, but here is the long way.

There are various ideas to know... all of them very, very old (think Ancient Greece), but here's the rundown.

#1. The sum of the angles of a triangle makes 180°

#2. Any line segment is also 180°. If you draw a point on the line segment, you can make "supplementary angles."... like ∠x + ∠y = 180°

#3. If you "bisect" something, you cut it in half. If a line segment bisects an angle, it cuts the angle in half.

*To undo that we multiply by 2.*

Step 1: What were we

*given?*- ABC is a triangle. That means:
- AC is a line segment
- We were told ∠BDC = 100°
- We were also told to draw point D on line AC
- line segment BD bisects ∠B

Step 2: Let's apply #2

We can do a little math to show ∠BDA = 80°

∠BDC + ∠BDA = AC = 180° (All line segments form a 180° angle)

∠BDC + ∠BDA = 180° (these angles are "supplementary")

∠BDC + ∠BDA = 180° (these angles are "supplementary")

100° + ∠BDA = 180°

∠BDA = 180° - 100° (this is the paperwork)

∠BDA = 80°

Step 3: Now we can look at Δ ABD.

∠A is still a right angle, so we can use #1

∠BDA = 80°

∠DAB = 90° (∠A is a right angle)

Therefore:

∠ABD = 180° - 90° - 80° (All angles of a triangle together make
180°)

∠ABD = 10°

Step 4: Remember that they told us that BD
bisects ∠B, so using
#3

∠B = ∠ABD * 2 = 20°

Step 5: We're getting close The last step is to solve
Δ ABC for ∠C

∠A = 90°

∠B = 20°

∠C = (180° - 90° - 20°)

Step 6: Finally, the solution!

**∠C = 70°**

## Comments

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