Math  /  Geometry

Question(From Unit 2, Lesson 3.)
4. Triangles ACDA C D and BCDB C D are isosceles. Angle BACB A C has a measure of 18 degrees and angle BDCB D C has a measure of 48 degrees.

The measure of angle ABDA B D is \qquad Show your work ADACand BDBCA D^{-} \cong A C^{-} \text {and } B D^{-} \cong B C^{-}

Studdy Solution

STEP 1

What is this asking? We need to find the angle ABDABD inside a diagram with two touching isosceles triangles, knowing some other angles. Watch out! Don't mix up the angles of the different triangles!
Isosceles triangles have two equal sides *and* two equal angles.

STEP 2

1. Find Angle *ADC*
2. Find Angle *DAC*
3. Find Angle *ACB* and Angle *ADB*
4. Find Angle *ABC*
5. Calculate Angle *ABD*

STEP 3

We know that the angles in a triangle add up to 180180^\circ.
In \triangle *BDC*, we know \angle *BDC* is 4848^\circ.
Since \triangle *BDC* is isosceles with BDBCBD \cong BC, we also know that \angle *DBC* == \angle *DCB*.
Let's call this angle *x*.

STEP 4

So, we have x+x+48=180x + x + 48^\circ = 180^\circ.
Simplifying, we get 2x+48=1802x + 48^\circ = 180^\circ.

STEP 5

Subtracting 4848^\circ from both sides gives us 2x=18048=1322x = 180^\circ - 48^\circ = 132^\circ.

STEP 6

Dividing both sides by 22 gives us x=1322=66x = \frac{132^\circ}{2} = 66^\circ.
So, \angle *DBC* == \angle *DCB* =66= 66^\circ.

STEP 7

We're given that \angle *BAC* is 1818^\circ.
Since \triangle *ACD* is isosceles with ADACAD \cong AC, we know \angle *ADC* == \angle *ACD*.
Let's call this angle *y*.

STEP 8

So, we have y+y+18=180y + y + 18^\circ = 180^\circ.
Simplifying, we get 2y+18=1802y + 18^\circ = 180^\circ.

STEP 9

Subtracting 1818^\circ from both sides gives us 2y=18018=1622y = 180^\circ - 18^\circ = 162^\circ.

STEP 10

Dividing both sides by 22 gives us y=1622=81y = \frac{162^\circ}{2} = 81^\circ.
So, \angle *ADC* == \angle *ACD* =81= 81^\circ.

STEP 11

We know \angle *ACD* is 8181^\circ and \angle *DCB* is 6666^\circ.
Adding these together gives us \angle *ACB*: 81+66=14781^\circ + 66^\circ = 147^\circ.

STEP 12

We also know \angle *ADC* is 8181^\circ and \angle *BDC* is 4848^\circ.
Adding these together gives us \angle *ADB*: 81+48=12981^\circ + 48^\circ = 129^\circ.

STEP 13

In \triangle *ABC*, we know \angle *BAC* =18= 18^\circ and \angle *ACB* =147= 147^\circ.
Since the angles in a triangle add up to 180180^\circ, we can find \angle *ABC*.

STEP 14

18+147+ABC=18018^\circ + 147^\circ + \angle ABC = 180^\circ, so 165+ABC=180165^\circ + \angle ABC = 180^\circ.

STEP 15

Subtracting 165165^\circ from both sides gives us ABC=180165=15\angle ABC = 180^\circ - 165^\circ = 15^\circ.

STEP 16

We know \angle *ABC* is 1515^\circ and \angle *DBC* is 6666^\circ.
Since \angle *ABD* and \angle *DBC* together make up \angle *ABC*, we have \angle *ABD* +66=15+ 66^\circ = 15^\circ.

STEP 17

Oops! It looks like we added the wrong angles earlier. \angle *ABC* is the larger angle, so we should have \angle *ABD* ++ \angle *DBC* == \angle *ABC*.

STEP 18

Let's call \angle *ABD* as *z*.
Then, z+66=15z + 66^\circ = 15^\circ.
Subtracting 6666^\circ from both sides gives z=1566=51z = 15^\circ - 66^\circ = -51^\circ.
Hmm, a negative angle?
That's not right!

STEP 19

Let's go back to \angle *ABC* =15= 15^\circ and \angle *DBC* =66= 66^\circ.
We made a mistake assuming *D* was inside triangle *ABC*.
It's actually outside.
So, \angle *ABD* == \angle *ABC* ++ \angle *DBC*.

STEP 20

Therefore, \angle *ABD* =15+66=81= 15^\circ + 66^\circ = 81^\circ.

STEP 21

The measure of angle ABDABD is 8181^\circ.

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