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ad-work [718]
3 years ago
11

12. In ∆PQR, if m∠P is 14 less than five times x, m∠Q is five less than x, and m∠R is nine less than twice x, find x and the mea

sure of each angle. 13. In ∆JKL, if m∠J is seven less than m∠L and m∠K is 21 less than twice m∠L, find the measure of each angle. 14. In ∆BCD, if BC ≅ BD , m∠B = 13x – 35, m∠C = 5x – 19, and m∠D = 2x + 14, find x and the measure of each angle. 15. In ∆WXY, if∠X ≅ ∠Y , WX = 9x – 11, XY = 4x + 1, and WY = 7x – 3, find x and the measure of each side.

Mathematics
1 answer:
pychu [463]3 years ago
3 0

Answer:

12)x=26.125°

∠P  = 115.625°

∠Q =21.125°

∠R=43.25°

13)x=52°

∠L=52°

∠J =45°

∠K =69°

14)x=11°

∠C = 36°

∠D =36°

∠B = 108°

15)x=7

WX = 52

WY=52

XY =29

Step-by-step explanation:

12)In ∆PQR

∠P is 14 less than five times x = 5x-15

∠Q is five less than x=x-5

∠R is nine less than twice x=2x-9

Angle sum property of triangle : The sum of all angles of triangle is 180°

So,5x-15+x-5+2x-9=180

8x-29=180

8x=180+29

x=\frac{180+29}{8}

x=26.125

So,∠P  = 5x-15 =5(26.125)-15=115.625°

∠Q =x-5=26.125-5=21.125°

∠R=2x-9=2(26.125)-9=43.25°

13)  In ∆JKL

Let the angle L be x

∠L=x

∠J is seven less than ∠L=x-7

∠K is 21 less than twice ∠L=2x-21

Angle sum property of triangle : The sum of all angles of triangle is 180°

So, x+x-7+2x-21=180°

4x-28=180°

4x=180+28

x=\frac{180+28}{4}

x=52

∠L=x=52°

∠J =x-7=52-7=45°

∠K =2x-21=2(45)-21=69°

14)In ∆BCD

BC ≅ BD(Given)

Opposite angles of equal sides are equal

So, ∠C =∠D

∠C = 5x – 19

∠D = 2x + 14

So, 5x-19=2x+14

3x=33

x=11

∠C = 5x – 19=5(11)-19=36°

∠D = 2x + 14=2(11)+14=36°

∠B = 13x – 35=13(11)-35=108°

15) In ∆WXY

∠X ≅ ∠Y(Given)

Opposite sides of equal angles are equal

WY=WX

WX = 9x – 11

WY = 7x – 3

So, 9x-11=7x-3

2x=14

x=7

So, WX = 9x – 11=9(7)-11=52

WY=52

XY = 4x + 1=4(7)+1=29

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__________________________________________________
Explanation:
_________________________________________________





                                5
--------------------------------------------
`                             right angle   |_ |
   `          (right triangle )                 |
         `                                               |    4 
             `                                           |
                      `    
       "c"                        `                  \        
 (hypotenuse)                          Starting point
                     
__________________________________________________
Since we have a "right triangle, we solve for "c"; using the 
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_______________________________________________________
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Plug these known values into our equation:
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 → 4² + 5² = c² ; 
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 → c =  √41 ; Use calculator;
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