Which is shown, an altitude, a median, or neither?
1 answer:
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#4) From the reference angle of 58° we can see that we have the side opposite to that angle as well as the hypotenuse. Recall that sin=opp/hyp so we are going to use sine to find that side
sin(58°) =
(multiply both sides by 19 to isolate x)
19 sin(58°) = x (plug into calculator)
16.1 = x
#5) From the reference angle of 56°, we see that we have the adjacent and the opposite sides. Remember that tan=opp/adj so we will use tangent to find x
tan(56°) =
(multiply both sides by 
)
(flip them so x is on the top)
[tex] \frac{12}{tan(56)} = x
8.1 = x
sin(58°) =
19 sin(58°) = x (plug into calculator)
16.1 = x
#5) From the reference angle of 56°, we see that we have the adjacent and the opposite sides. Remember that tan=opp/adj so we will use tangent to find x
tan(56°) =
[tex] \frac{12}{tan(56)} = x
8.1 = x
The radius of the circle is 3 cm.
<u>Step-by-step explanation:</u>
Refer the attached diagram, the circle with centre O. In that given, AB is tangent given as 4 cm and distance of point from the circle, OA = 5 cm
As AB is tangent, OB (radius of circle) is perpendicular to AB (tangent at any point of circle). Therefore the angle of OBA is 90 degree.
Also, triangle OAB is a right angled triangle (refer attached diagram). By using Pythagoras theorem in right angled triangle,
Substitute the given values in the above expression, we get
Taking square root on both side, we get
Radius of the circle, OB = 3 cm
