Suppose a triangle has vertices AA, BB and CC and has the following measurements:¯¯¯¯¯¯¯¯AB=23.5 cm ¯¯¯¯¯¯¯¯AC=87.2 cm ¯¯¯¯¯¯¯¯BC=64.8 cm Start by drawing a diagram of this triangle and labeling the known values.

What is the degree measure of ∠C?
m∠C=  °   

What is the degree measure of ∠A?
m∠A=  °   

What is the degree measure of ∠B?
m∠B=  °   

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answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)