Question

In a quadrilateral ABCD,the diagonals AC and BD intersect at O.If <DAB = <BCD = 60°,DO = 4cm and AO= 5cm.Find the value of AC+BC.

Answer 1

AC + BD = 18.66 cm

Explanation:

given data

$\angle$ DAB = $\angle$ BCD = 60°

DO = 4 cm

AO= 5 cm

to find out

AC+BC

solution

we know when ∠A = ∠C = 60°

so than there will be ∠B = ∠D = 120° as per symmetry

and it will become also a parallelogram  

so here

AC = 2 OA = 10    ....................1

BD = 2 OD = 8 cm    .......................2

and

we take here triangle ABC

so here sin(120) = $\frac{BC}{AC}$  

put here value and we get BC

sin(120) = $\frac{BC}{10}$  

BC = 8.66

so that AC + BC will be

AC + BD = 10 + 8.66

AC + BD = 18.66 cm

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quadrilateral

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