Answer:
13.5 cm
Step-by-step explanation:
if 5 cm is 2 km, then 10 cm is 4 km. To find one km you do half of 5 cm. which is 3.5 cm. So 10+3.5=13.5
Keep in mind that, when the logarithm base is not explicitly written, base 10 is assumed, therefore,
We know that : Sum of Angles in a Triangle is equal to : 180°
⇒ In ΔRST, The Sum of Angles ∠R , ∠S , ∠T should be equal to 180°
⇒ m∠R + m∠S + m∠T = 180°
⇒ (2x + 10)° + (2x + 25)° + (x - 5)° = 180°
⇒ (2x + 2x + x) + (10° + 25° - 5°) = 180°
⇒ 5x + 30° = 180°
⇒ 5x = 180° - 30°
⇒ 5x = 150°
⇒ x = 30°
Answer:
Conclusion:
Step-by-step explanation:
Given
We know that the diagonals of a parallelogram bisect each other.
Therefore,
Given RT = x and TP = 5x-28, so
x = 5x-28
5x = x+28
5x-x = 28
4x = 28
divide boh sides by 4
4x/4 = 28/4
x = 7
Thus, the value of x = 7
Similarly,
QT = TS
Given QT = 5y and TS = 2y+12, so
5y = 2y+12
5y-2y = 12
3y = 12
divide both sides by 3
3y/3 = 12/3
y = 4
Thus, the value of y = 4
Conclusion:
Answer:
∠BAD=20°20'
∠ADB=34°90'
Step-by-step explanation:
AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.
Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus
∠BOD+∠BDO+∠DBO=180°
∠BDO+∠DBO=180°-110°20'=69°80'
∠BDO=∠DBO=34°90'
So ∠ADB=34°90'
Angles BOD and BOA are supplementary (add up to 180°), so
∠BOA=180°-110°20'=69°80'
In right triangle ABO,
∠ABO+∠BOA+∠OAB=180°
90°+69°80'+∠OAB=180°
∠OAB=180°-90°-69°80'
∠OAB=20°20'
So, ∠BAD=20°20'