(-7,-3) (-1,0) (9,5) and (13,7)
Answer:
<h2>b = 15°</h2>
Step-by-step explanation:
If Pq = RQ then ΔPQR is the isosceles triangle. The angles QPR and PRQ have the same measures.
We know: The sum of the measures of the angeles in the triangle is equal 180°. Therefore we have the equation:
m∠QPR + m∠PRQ + m∠RQP = 180°
We have
m∠QPR = m∠PRQ and m∠RQP = 60°
Therefore
2(m∠QPR) + 60° = 180° <em>subtract 60° from both sides</em>
2(m∠QPR) = 120° <em>divide both sides by 2</em>
m∠QPR = 60° and m∠PRQ = 60°
Therefore ΔPRQ is equaliteral.
ΔPSR is isosceles. Therefore ∠SPR and ∠PRS are congruent. Therefore
m∠SPR = m∠PRS
In ΔAPS we have:
m∠SPR + m∠PRS + m∠RSP = 180°
2(m∠SPR) + 90° = 180° <em>subtract 90° from both sides</em>
2(m∠SPR) = 90° <em>divide both sides by 2</em>
m∠SPR = 45° and m∠PRS = 45°
m∠PRQ = m∠PRS + b
Susbtitute:
60° = 45° + b <em>subtract 45° from both sides</em>
15° = b
Answer:
The wide of river is 
Step-by-step explanation:
In this problem we know that
Triangles DEG and DEF are congruent by ASA (Angle-Side-Angle) Congruence
therefore
EG=EF
Answer:
your answer is D.
Step-by-step explanation:
Using sine will give you 3*square root of 3
Answer:
the area of the circle: 25π
Step-by-step explanation:
45/4 π = 11.25π
9/10π = 0.9π
11.25π / 0.9π = x / 2π .... 360° = 2π
x = (11.25π x 2π) / 0.9π = 25π
