Question

in ∆ABE, sin 60°=AE/AB
√3/2=AE/16
=>AE=8√3 cm
and cot 60°=BE/AE
1/√3=BE/8√3
=>BE= 8 cm
EC=BC - BE
=23-8=15 cm
in rt ∆ AEC,
by Pythagoras theorem,
AC sq. = AE sq.+EC sq.
=(8√3) sq. + 15 sq.
=192+225
417
=>AC=√417
=20.42
.. now I am noticing.. question explain karte karte answer nikal diye

​

Answer 1

Answer:

lol!!!

answer pataa hi sis tho question pucha hi kyu?

Answer 2

Dekh . . iss bhi easy method..

∆ABE.. m B=60 ° , E=90°,

Toh ,.. . A=30°,..

Toh Yeh Ek 30,60 and 90 ka ∆ HAI. .

Toh , side opp to 30° =1/2 ×hypo..

BE=1/2×16

BE= 8...

Side Opp to 60° = √3/2 × hypo. .

AE=8√3

BC=BE+EC

EC=.. 15cm. .

∆AEC...

AE²+EC²=AC²

64*3 + 15²=

225+ 192 =