Question

PLEASE HELP ABCD is a trapezium. Angle ADC = Angle BCD = 90. Angle BAD = 58. AB = 4.7cm and BC = 7.3cm. a) Show that AD = 9.8cm correct to 1 decimal place. b) Find the size of Angle ACB?

Answer 1

see below

Explanation:

Note from the diagram that length AD is AP + PD. We know the length PD, this is the same as BC.

Finding length AP.

`AP=4.7cos(58^o)=2.491` cm ( 3 .d.p.)

So length of AD:

`AD= AP+PD=2.491+7.3=9.791=9.8` ( 1 .d.p.)

Angle ACB

First we need to find the length of AC. We can do this by finding the length CD. This is the same length as PB:

Length PB:

`PB=CD=4.7sin(58^o)=3.99` ( 2.d.p.)

`AC^2=AD^2+CD^2`

`AC^2=(9.8)^2+(3.99)^2=111.9601`

`AC=sqrt(111.9601)=10.58` ( 2 .d.p.)

Now we know all the sides of triangle ABC, we can use the Cosine Rule to find angle ACB.

`cos(C)= (a^2+b^2-c^2)/(2ab)`

`cos(C)=((7.3)^2+(10.58)^2-(4.7)^2)/(2(7.3)(10.58))=0.926641116`

`/_ACB=cos^-1(cos(C))=cos^-1(0.926641)=color(blue)(22.08^o)`