Question

3.0

Answer 1

Answer:

Solution given:

AD=1.8

AC=3.0

we have

∆ADC is a right angled triangle

Cos A=b/h

<A=cos-¹ (1.8/3)

<A=53.13°

again

∆ABC is a right angled triangle

Tan A=p/b=BC/AC

Tan 53.13×3=BC

BC=4

NOW

Area of triangle =½(AC*BC)=½(4×3)=6 square units is a required area.