1. Make a reasonable sketch.
draw radii to A and B
Triangle ABO is isosceles with sides 52 , 52 , and 96
By the cosine law we can find cosA
52^2 = 52^2 + 96^2 - 2(96)(52)cosA
cosA = 96^2/(2*52*96) = 12/13
In triangle AMO
MO^2 = 33^2 + 52^2 - 2(33)(52)cosA
= 1089 + 2704 - 3432(12/13) = 625
OM = √625 = 25
Give me a bit of time for #2
1. Ab is a chord of a circle with center o and radius 52 cm . point m divides the chord ab such that am = 63 cm and mb=33 cm find om
2. A circle is inscribed in a triangle whose sides are 10, 10 and 12 units . a second smaller circle is inscribed tangent to the first circle and to the equal sides of the triangle. Find the radius of the second triangle.
Help me guys thnk you its very much appreciated
2 answers
#2
Notice the centre of the inscribed circle must lie on the bisectors of the angles.
We have an isosceles triangle, with sides 10,10 and 12
If we place the triangle on the x-y grid, we can have the base from (-6,0) to (6,0) and the third vertex on the y-axis.
The y-axis becomes the altitude. Notice that in the right-angled triangles, we have a base of 6 and a hypotenuse of 10. You should recognize double the 3-4-5 triangle, so the height is 8 and the third vertex is (0,8)
label (-6,0) as B and A is (0,8)
slope of AB = (8-0)/(0-(-6)) = 4/3
An important property in trig is that the slope of a line is equal to the tangent of the angle that line makes with the x-axis
so we can find the angle at B
tan B = 4/3
B = 53.1301....
Since we need the angle of the bisector of angle B
that angle would be 53.1301../2
and the slope of that bisector is
tan (53.1301.../2) = 1/2 exactly
(are you surprised ?? We could show that it is exactly 1/2 by using Tan (2Ø) = 2tanØ/(1-tan^2 Ø) )
So the equation of the bisector from B is
y = (1/2)x + b, with (-6,0) on it
0 = (1/2)(-6) + b
b = 3
ahhh, so we have the centre of the large circle at (0,3) , the y-intercept
So the big circle cuts the y-axis at (0,6) , its diameter being 6
Label that point P , draw a horizontal line at P cut cut AB at Q
Because of parallel lines, angle AQP = angle B
and the slope of the angle bisector of angle AQP is also 1/2
How about the equation of the angle bisector of angle Q ??
We know its equation must be y = (1/2)x + k , ( I don't want to use b again)
So we need Q
equation of AB: y = (4/3)x + 8
at Q, y = 6
6 = (4/3)x + 8
-2 = (4/3)x
x = -2(3/4) = -3/2
So we have Q as (-3/2,6)
Q must lie on the angle bisector of angle Q, which is y = (1/2)x + k
ALL WE NEED IS K, AND WE ARE BASICALLY DONE
y = (1/2)x + k
6 = (1/2)(-3/2) + k
k = 6 + 3/4
so the centre of the little circle is at (0 , 6 3/4 )
but the large circle covered 6 of that distance
So the radius of the little circle is 3/4
(you might want to print out that solution for easier reading.)
The solution looks very long and complicated, but the essential work is really quite short, most of the solution consists of explanations.
Perhaps one of the other tutors might have a quicker way, sometimes it is easy to miss the obvious once your mind locks in on a method.
Notice the centre of the inscribed circle must lie on the bisectors of the angles.
We have an isosceles triangle, with sides 10,10 and 12
If we place the triangle on the x-y grid, we can have the base from (-6,0) to (6,0) and the third vertex on the y-axis.
The y-axis becomes the altitude. Notice that in the right-angled triangles, we have a base of 6 and a hypotenuse of 10. You should recognize double the 3-4-5 triangle, so the height is 8 and the third vertex is (0,8)
label (-6,0) as B and A is (0,8)
slope of AB = (8-0)/(0-(-6)) = 4/3
An important property in trig is that the slope of a line is equal to the tangent of the angle that line makes with the x-axis
so we can find the angle at B
tan B = 4/3
B = 53.1301....
Since we need the angle of the bisector of angle B
that angle would be 53.1301../2
and the slope of that bisector is
tan (53.1301.../2) = 1/2 exactly
(are you surprised ?? We could show that it is exactly 1/2 by using Tan (2Ø) = 2tanØ/(1-tan^2 Ø) )
So the equation of the bisector from B is
y = (1/2)x + b, with (-6,0) on it
0 = (1/2)(-6) + b
b = 3
ahhh, so we have the centre of the large circle at (0,3) , the y-intercept
So the big circle cuts the y-axis at (0,6) , its diameter being 6
Label that point P , draw a horizontal line at P cut cut AB at Q
Because of parallel lines, angle AQP = angle B
and the slope of the angle bisector of angle AQP is also 1/2
How about the equation of the angle bisector of angle Q ??
We know its equation must be y = (1/2)x + k , ( I don't want to use b again)
So we need Q
equation of AB: y = (4/3)x + 8
at Q, y = 6
6 = (4/3)x + 8
-2 = (4/3)x
x = -2(3/4) = -3/2
So we have Q as (-3/2,6)
Q must lie on the angle bisector of angle Q, which is y = (1/2)x + k
ALL WE NEED IS K, AND WE ARE BASICALLY DONE
y = (1/2)x + k
6 = (1/2)(-3/2) + k
k = 6 + 3/4
so the centre of the little circle is at (0 , 6 3/4 )
but the large circle covered 6 of that distance
So the radius of the little circle is 3/4
(you might want to print out that solution for easier reading.)
The solution looks very long and complicated, but the essential work is really quite short, most of the solution consists of explanations.
Perhaps one of the other tutors might have a quicker way, sometimes it is easy to miss the obvious once your mind locks in on a method.
1 answer