Question

22. ABCD is a trapezium in which AB || CD, AB = 14 cm, BC = AD = 5 cm and DC = 8 cm. Find the value of
3 sin B-4 cos B.
Please solve step by step...​

Answer 1

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Let AB=11 cm, CD=6 cm and AB is parellel to CD.

Let distance between foot of perpendicular h and A

is X.

So distance between B and foot of 2nd perpendicular from C=11–6-x=5-x.

There are 2 rt. angled triangles.

Applying Pythogorous theorem,

For one triangle,

h^2=3^2-x^2—————(1)

For 2nd triangle,

h^2=4^2-(5-x)^2————-(2)

From (1)&(2), x=1.8 cm and

required h=2.4 cm

Answer 2

Answer:

given that :-ABCD is a trapezium in which AB || CD, AB = 14 cm, BC = AD = 5 cm and DC = 8 cm.

To find 3sinB - 4cosB

Step-by-step explanation:

DM⊥AB and CN⊥BC

AM=NB

AM+NB=14-8

2NB=6

NB=3 cm

CN^2=BC^2-NB^2

CN^2=5X5-3X3

CN^2=25-9

CN=4CM

sinB=4/5

cosB=3/5

3sinB-4cosB= 3x(4/5)-4x(3/5)

=12/5-12/5

=0