Answer:
Step-by-step explanation:
<h3>Q13</h3>
The greater the angle the greater the opposite side
<u>Sides in ascending order:</u>
- AB = 17, AC = 18, BC = 21
<u>Angles in same order</u>
<h3>Q14</h3>
<u>As above, sides in ascending order:</u>
- AB = 15, AC = 16, BC = 17
<u>Angles in same order</u>
<h3>Q15</h3>
<u>Exterior angle equals to sum of non-adjacent interior angles</u>
- 142° = x + 66°
- x = 142° - 66°
- x = 76°
<h3>Q16</h3>
<u>Same subject and isosceles triangle:</u>
- x + x = 158°
- 2x = 158°
- x = 79°
<h3>Q17</h3>
<u>Same subject</u>
- m∠QSR = m∠QPS + m∠PQS
- 2x = x + m∠PQS
- m∠PQS = 2x - x
- m∠PQS = x
ΔPQS has two angles with the measure of x, hence their opposite sides are congruent and the triangle is isosceles
Answer:
(−∞,∞). no solution
Step-by-step explanation:
Answer:
Took a lot of thinking but y= 840
Step-by-step explanation:
Step-by-step explanation:
2pir/2 + 2 + 2(14) + 2piR/2
= pir + 2 + 28 + piR
r = [(5+5)-2]/2
r = 4
Perimeter = 3.14(4) + 30 +3.14(5)
= ........... meters
The cofunction of cos is sin(90-x)
90 degrees is equal to PI/2
The cofunction becomes sin(PI/2 - 2PI/9)
Rewrite both fractions to have a common denominator:
PI/2 = 9PI/18
2PI/9 = 4PI/18
Now you have sin(9PI/18 - 4PI/18)
Simplify:
Sin(5PI/18)