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Are you able to solve these problems?
1.

2.

3.

 Apr 8, 2024
 #1
avatar+128794 
+1

1.

 

tan x + sec x =  2

 

sinx / cos x  + 1/cos x  = 2

 

sinx + 1   = 2cos x           square bith sides

 

sin^2 x + 2sin x + 1  = 4cos^2 x

 

sin^2 x + 2sin x + 1 =  4 (1-sin^2 x)

 

sin^2 x + 2sin x + 1 =  4 - 4sin^2 x

 

5sin^2 x + 2sin x - 3 =   0

 

(5sin x -3  )(sinx + 1)   = 0 

 

5sin x - 3  =  0                      sin x + 1  =  0

 

5sinx  = 3                              sin x  = -1

 

sin x =  3/5                                x = -pi/2     reject

 

cos x =   sqrt [ 5^2 - 3^2 ] / 5  =  sqrt [ 16 ] / 5  =   4 / 5

 

 

cool cool cool

 Apr 8, 2024
 #2
avatar+128794 
+1

2.

 

Call the point where the diagonals meet =  O

 

Triangle SOR  similar to triangle XOY

 

XY  / SR  = 5/28

 

So...height of triangle XOY  to height of triangle SOR = 5/28

 

Then there are 5 + 28  =  33  equal parts  in 1/2 height of the trapezoid

 

And triangle POQ similar to  triangle SOR

 

And height of POQ to height of SOR =    [ 5 +33] / 28  =  38/28   = 19/14

 

So

 

PQ  =  (19/14)SR=  (19/14) * 28 =   38 

 

cool cool cool

 Apr 8, 2024
 #3
avatar+128794 
+1

3.   Angle  B  = 180 - 55 -26  = 99°

 

10 / sin 99 = BC / sin  55

 

BC =  10sin55 / sin 99    ≈  8.3

 

 

cool cool cool

 Apr 8, 2024

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