Solve: 4(2x - 1) - 3 = 9(x - 4) -29 32 13 29
1 answer:
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Answer:
a. 360.323 m
b.
Step-by-step explanation:
Distance covered from A to B = speed x time
= 110 x 2.8
= 308 m
Distance covered from B to C = speed x time
= 110 x 1.7
= 187 m
The sum of angles at B = +
=
a. The distance of the plane from it starting point to C can be determined by applying the cosine rule.
=
+
- 2ac Cos B
Sot hat;
=
+
- 2(187 x 308) Cos
But, Cos = 0
So that,
=
+
= 34969 + 94864
= 129833
b =
= 360.323
The distance from A to C is 360.323 m.
b. Applying the sine rule;
=
=
=
Sin A =
= 0.5190
⇒ A = 0.5190
=
The bearing of the plane from its original location = +
=
----------------------------------------------------------------------------------------------------
Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
----------------------------------------------------------------------------------------------------
In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
----------------------------------------------------------------------------------------------------
by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°
Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
----------------------------------------------------------------------------------------------------
In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
----------------------------------------------------------------------------------------------------
by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°