In a rhombus ABCD, m∠A = 31°. Point O is a point of intersection of diagonals. Find the measures of the angles of triangle ΔBOC.
I got 15.5 degrees for angle OCB and thats right all I need is the angles for OBC and BOC
<BOC = 90
<OBC = 74.5
note that ΔBOC is a right triangle, since the diagonals are perpendicular.
To find the measures of angles OBC and BOC in triangle ΔBOC, we can make use of the properties of rhombus and the fact that the sum of the angles in a triangle is 180°.
In a rhombus, the diagonals intersect at 90° angles. Therefore, angle BOC is a right angle (90°).
Now, let's consider angle OBC. We know that angle BOC is 90°, and angle AOB is also 90° since the diagonals of a rhombus bisect each other at right angles. Therefore, we can see that triangle ΔBOC is actually a right-angled triangle.
Since angles in a right-angled triangle sum up to 180°, we can calculate angle OBC using the formula:
angle OBC = 180° - angle BOC
Plugging in the value of angle BOC as 90°, we get:
angle OBC = 180° - 90°
angle OBC = 90°
So, angle OBC is 90°.
Now, to find angle BOC, we can again use the fact that triangle ΔBOC is a right-angled triangle, which means the sum of the angles in this triangle is 180°. Therefore:
angle BOC = 180° - angle BCO - angle OBC
Plugging in the values of angle BCO as 31° and angle OBC as 90°, we get:
angle BOC = 180° - 31° - 90°
angle BOC = 180° - 121°
angle BOC = 59°
So, angle BOC is 59°.
In summary, the measures of the angles in triangle ΔBOC are:
Angle OBC = 90°
Angle BOC = 59°
m∠BOC = 90 degrees
, m∠OCB = 15.5 degrees
, m∠OBC = 74.5 degrees
why does everyone keep getting BOC=90?
that is wrong.