Two sides of a parallelogram measure 60 centimeters and 40 centimeters. If one angle of the parallelogram measures 132, find the length of each diagonal.

By cosine rule, the length of the longer diagonal is sqrt(60^2 + 40^2 - 2 x 60 x 40 cos 132) = sqrt(8,411.83) = 91.7 cm

The other angle of the parallelogram is 180 - 132 = 48
The length of the shorter diagonal is sqrt(60^2 + 40^2 - 2 x 60 x 40 cos 48) = sqrt(1,988.17) = 44.6 cm

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parmesanchilliwack parmesanchilliwack · Answer 2 · 2019-03-20T11:14:29+01:00

Answer: The length diagonals of the given parallelogram are approximately 91.716 cm and 44.589 cm.

Step-by-step explanation:

Since, a parallelogram has two pairs of congruent opposite angles

and two pairs of congruent sides,

Let ABCD is a parallelogram ( shown below)

In which AB = CD = 60 meters and BC = DA = 40 meters

∠ ABC = ∠ADC = 132° ⇒ ∠DAB = ∠DCB = 1/2(360 - 2×132)= 48°

We have to find : AC and BD.

In triangle ABC,

By the cosine law,

[tex]AC^2=AB^2+BC^2-2\times AB\times BC\times cos(132^{\circ})[/tex]

[tex]AC^2=60^2+40^2-2\times 60\times 40\times-0.66913060635[/tex]

[tex]AC^2=3600+1600+3211.82691052=8411.82691052[/tex]

[tex]AC=91.7160123\approx 91.716\text{ cm}[/tex]

Again, in triangle ABD,

[tex]BD^2=AB^2+AD^2-2\times AB\times AD\times cos(48^{\circ})[/tex]

[tex]BD^2=60^2+40^2-2\times 60\times 40\times 0.66913060635[/tex]

[tex]BD^2=3600+1600-3211.82691052=1988.17308948[/tex]

[tex]BD=44.5889346\approx 44.589\text{ cm}[/tex]

hence, the diagonals of the given parallelogram are approximately 91.716 cm and 44.589 cm.