UNSOLVEABLE . I will Mark Brainliest Given square ABCD. Two isosceles triangles ABP and BCQ are constructed with bases AB and BC . Each of these triangles has vertex angle of 80°. Point P lies in the interior of the square, while point Q lies outside of the square. Find the angle measure between PQ and BC

please get this correct no one else has been able to solve

I'm Not Sure

Step-by-step explanation:

Transcript. Ex7.2, 5 ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD. Given: ΔABC is isosceles So, AB = AC ΔDBC is isosceles So, DB = DC To prove: ∠ ABD

Step-by-step explanation:

answer: 91

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."