Consider the diagram and the proof below.
Given: In △ABC, AD ⊥ BC
Prove: StartFraction sine (...
Mathematics, 24.06.2020 23:01, vanessacox45
Consider the diagram and the proof below.
Given: In △ABC, AD ⊥ BC
Prove: StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction
Triangle A B C is shown. A perpendicular bisector is drawn from point A to point D on side C B forming a right angle. The length of A D is h, the length of C B is a, the length of C A is b, and the length of A B is c.
A 2-column table has 7 rows. The first column is labeled Statement with entries In triangle A B C line segment A D is perpendicular to line segment B C, In triangle A D B sine (uppercase B) = StartFraction h Over c EndFraction, c sine (uppercase B) = h, In triangle A C D, sine (uppercase C) = StartFraction h Over b EndFraction, b sine (uppercase C) = h, question mark, StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction. The second column is labeled Reason with entries given, definition of sine, multiplication property of equality, definition of sine, multiplication property of equality, substitution, and division property of equality.
What is the missing statement in Step 6?
b = c
StartFraction h Over b EndFraction = StartFraction h Over c EndFraction
csin(B) = bsin(C)
bsin(B) = csin(C)
Answers: 3
Mathematics, 21.06.2019 18:20, Gigglygoose4181
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Mathematics, 21.06.2019 20:30, nosugh
If m∠abc = 70°, what is m∠abd? justify your reasoning. using the addition property of equality, 40 + 70 = 110, so m∠abd = 110°. using the subtraction property of equality, 70 − 30 = 40, so m∠abd = 30°. using the angle addition postulate, 40 + m∠abd = 70. so, m∠abd = 30° using the subtraction property of equality. using the angle addition postulate, 40 + 70 = m∠abd. so, m∠abd = 110° using the addition property of equality.
Answers: 2
Mathematics, 21.06.2019 21:00, memoryofdale
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