Question

In trapezoid ABCD the lengths of the bases AD and BC are 7 and 5 respectively, and the length of diagonal AC is 6. The diagonals are perpendicular. Find the measure of ∠BDA.

Answer 1

Answer:

The measure of ∠BDA  is 30 Degrees.

Step-by-step explanation:

Lets construct a trapezium with the given data

If O is the point of intersection of the diagonals, ∆ AOD ~ ∆COB. Then

$\frac{ AO}{AD} = \frac{CO}{CB}$

On substituting the given values

$\frac{ AO}{7} = \frac{CO}{5}$

Also we know that AC = 6 . That is

AO + OC = 6 or  OC = 6 - AO  

substituting for CO and cross multiply

5·AO = (6-AO)·7

5AO  = 42 - 7AO

12AO = 42

AO  = $\frac{42}{12}$ = 3.5

All of the angles at E are right angles, so $\angle BDA$ has the trigonometric ratio

$sin(\angle BDA) = \frac{AO}{AD}$ = $\frac{3.5}{7}$  = 0.5 = $\frac{1}{2}$

∠BDA = arcsin(1/2) = 30°