Question

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.

Answer 1

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

Answer 2

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,

$AC^2=AB^2+BC^2-2\times AB\times BC\times cos(132^{\circ})$

$AC^2=60^2+40^2-2\times 60\times 40\times-0.66913060635$

$AC^2=3600+1600+3211.82691052=8411.82691052$

$AC=91.7160123\approx 91.716\text{ cm}$

Again, in triangle ABD,

$BD^2=AB^2+AD^2-2\times AB\times AD\times cos(48^{\circ})$

$BD^2=60^2+40^2-2\times 60\times 40\times 0.66913060635$

$BD^2=3600+1600-3211.82691052=1988.17308948$

$BD=44.5889346\approx 44.589\text{ cm}$

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