In a cyclic quadrilateral ABCD twice angle BAD =three times angle DCB. Calculate angle BAD
BAD+BCD=180
2BAD=3 BAD,so BCD =2/3BAD
5/3BAD=180
BAD=108
To solve this problem, we'll first use the given information to set up an equation. Let x be the measure of angle BAD and y be the measure of angle DCB.
Given: Twice angle BAD = Three times angle DCB
We can write this equation as:
2x = 3y
Next, we use the fact that the sum of the angles in a cyclic quadrilateral is equal to 360 degrees.
In quadrilateral ABCD, the sum of angles BAC, BCD, and BAD is equal to 360 degrees.
Hence, we have:
(angle BAD) + (angle DCB) + (angle BAC) + (angle BCD) = 360 degrees
Substituting the values of x and y into the equation, we get:
x + y + (angle BAC) + (angle BCD) = 360 degrees
Since the angles in a cyclic quadrilateral are supplementary (opposite angles add up to 180 degrees), we can rewrite the equation as:
x + y + (180 degrees - x) + (180 degrees - y) = 360 degrees
Simplifying the equation, we obtain:
x + y + 180 degrees - x + 180 degrees - y = 360 degrees
Combining like terms, we get:
2y + 360 degrees = 360 degrees
Now, let's solve for y:
2y = 0 degrees
y = 0 degrees / 2
y = 0 degrees
Since y = 0 degrees, we can substitute this value back into the original equation to solve for x:
2x = 3y
2x = 3(0 degrees)
2x = 0 degrees
x = 0 degrees / 2
x = 0 degrees
Therefore, the measure of angle BAD (x) in this cyclic quadrilateral is 0 degrees.
The key here is to recognize that opposite angles are supplementary. So,
BAD+BCD = 180
2BAD=3BCD, so BCD = 2/3 BAD
5/3 BAD = 180
BAD = 108