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# Geometry!

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Are you able to solve these problems?
1.

2.

3.

Apr 8, 2024

#1
+128794
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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

Apr 8, 2024
#2
+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

Apr 8, 2024
#3
+128794
+1

3.   Angle  B  = 180 - 55 -26  = 99°

10 / sin 99 = BC / sin  55

BC =  10sin55 / sin 99    ≈  8.3

Apr 8, 2024