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# help with hard problems!

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1. In a triangle, the measure of the first angle is three times the measure of the second angle. The measure of the third angle is 80° more than the measure of the second angle. Use the fact that the sum of the measures of the three angles of a triangle is 180° to find the measure of each angle.

2. A circle with radius r is inscribed into a right triangle. Find the perimeter of the triangle if:the length of the hypotenuse is 24 cm, and r=4 cm;

3. Right triangle ABC has legs AC and BC of lengths 16mi and 24mi accordingly. Find the distance of point B from the line that contains the median AM.

4. Given triangle ABC, AD is perpendicular to BC such that DC = 2 and BD = 3.  If angle BAC = 45 degrees, then find the area of triangle ABC.

5. CPhill has answered all your others but I expect you can do the last one by yourself. - Melody.

Feb 13, 2020
edited by Melody  Feb 14, 2020

#1
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1. In a triangle, the measure of the first angle is three times the measure of the second angle. The measure of the third angle is 80° more than the measure of the second angle. Use the fact that the sum of the measures of the three angles of a triangle is 180° to find the measure of each angle.

Let the second angle  = A

The first angle  = 3A

And the remaining angle   = A + 80

So

A + 3A  + A +  80   = 180           simplify

5A + 80  = 180     subtract   80 from both sides

5A   = 100   divide  both sides  by  5

A  = 20

Second angle =  60

Remaining angle = 100   Feb 13, 2020
#5
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1st angle = 60

2nd .......= 20

3rd........= 100 Guest Feb 14, 2020
#2
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Let C  =(0,0)

Let A = (0,16)

Let B = (24,0)

The median, M  =  (12,0)

AM  has the slope     (16-0)/ ( 0-12)  = -16/12=   -4/3

And the equation  of  AM  is

y =(-4/3)x  + 16

In stadard form,  Ax + By  - C  = 0,   the equation is

3y = -4x + 48

4x + 3y  - 48   = 0

The distance  from B = (24,0)    to  this line is

l  4 (24)  + 3(0)  -  48   l                     l  48  l

___________________    =            ----------   =     48/5 mi   =     9.6 mi

sqrt  ( 4^2 + 3^2 ]                                5   Feb 13, 2020
#3
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4. Given triangle ABC, AD is perpendicular to BC such that DC = 2 and BD = 3.  If angle BAC = 45 degrees, then find the area of triangle ABC.

We have something  like this   :

A

B          3             D              2                 C

Using the Pythagorean Theorem :

AC^2 =  4  +  AD^2       subtract these

AB^2  - AC^2  =  5

AB^2  = 5 + AC^2

AB =  sqrt  (5 + AC^2)

Using the Law of Cosines

BC^2  = AB^2  + AC^2  - 2 (AB * AC)cos (45°)

Let x =   AC

5^2 = (5 + x^2)  + x^2  -  2* [ sqrt (5 + x^2) * x] (sqrt (2) /2 )

25 = 5 + x^2 + x^2  - sqrt (2)  [ sqrt (5 + x^2) * x ]

Solving this for the positive value of x  gives  x =  2sqrt (10)  =  sqrt (40) =  AC

And AB = sqrt (5 + AC^2) = sqrt (5 + 40) = sqrt (45)

So....the area of triangle  ABC =

(1/2) AC * AB  sin (45°) =

(1/2) (sqrt (40) * sqrt (45) * [ sqrt (2) / 2 ]  =

(1/4) * sqrt (3600)  =

(1/4) * 60  =

15 units^2   Feb 14, 2020
edited by CPhill  Feb 14, 2020
#4
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2. A circle with radius r is inscribed into a right triangle. Find the perimeter of the triangle if:the length of the hypotenuse is 24 cm, and r=4 cm;

See the following image : Let

AE = 4

EC = a

CG = a

BG = b

BF = b

FA = 4

So  the sides are

a + b  = 24

b + 4

4 + a

And

a + b =  24    →   b = 24  - a

b + 4   →    28 - a

4 + a

Using the Pythagorean Theorem :

(4 + a)^2  + ( 28 - a)^2  =  24^2

Solving this for a produces

a =  4 ( 3 + sqrt (2) )   =  12  + 4sqrt (2)

4 + a =  16 + 4sqt (2)

b = 24 - a  =  24 - (12 + 4 sqrt (2) )  =  12 - 4sqrt (2)

And b + 4 =  16 - 4sqrt (2)

So....the perimeter  =

24 + (16 - 4sqrt (2)) + (16 + 4sqrt (2) )  =

24  + 32

56 units   Feb 14, 2020
edited by CPhill  Feb 14, 2020