Answer:
Step-by-step explanation:
Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.
The tangent passing through (6,2) would have equaiton of the form
for suitable m.
Because this line is tangent, distance of centre of circle namely origin is the radius 1.
i.e.
the coordinates are the intersection of the tangent with the circle
They are (-0.162, 0.987) and (0.462,-0.887)
Answer:
15 m
explanation:
Let AB and CD be the pole and tower respectively. Let CD=h Then ∠DAC=60 ∘ and∠DBE=30 ∘ Now CA CD =tan60 ∘ = 3 ∴CD= 3 CA⇒ 3 h =CA Again BE DE =tan30 ∘ = 3 h ∴(h−10)= 3 BE = 3 CA ...[∵BE=CA] = 3 h/ 3 = 3 h ⇒3h−30=h⇒2h=30⇒h=15
Hence, height of the power =15 m
Answer:
14.25
Step-by-step explanation:
95 x 0.85 = 80.75
95-80.75 = 14.25
Answer:
Step-by-step explanation: