1. A=33°, a=20, b=13
using sine rule:
a/sin A=b/sin B=c/sin C
the value of B will be
20/sin 33=13/sin B
sin B=(13sin33)/20
sin B=0.3540
B=20.7°
to get C we proceed as follows:
a=33°, b=20.7° hence C will be:
C=180-(33+20.7)
C=126.3°
to get cwe use cosine rule:
c²=a²+b²-2abcos C
substituting the values we get:
c²=20²+13²-2×20×13cos 126.3
c²=876.85
c=√876.85
c=29.6
The answer is:
B=20.7°, C=126.3°, c~29.6
2] The measure measurements of C=30°, a=32, c=16 determine the measurements of one triangle whose sides can be obtained using cosine rule and sine rule. Using these rules we can only obtained the dimensions of one triangle given the measurements provided.
sin A/32=sin 30/16
16sin A=32*1/2
16sin A=16
sin A=1, and thus A=90° only.
This implies that only 1 triangle can be formed.
3. A=59°, a=13, b=14. Using sine rule:
a/ sin A=b/sin B=c/sin C
thus:
13/sin 59=14/sin B
sin B=0.92310
B~67.375 or 112.583
This implies that there are two possibletriangles which can be formed.
from the first triangle:
a=13, b=14, A=59°, B=67.375°, C=180-(A+B)=53.625°
Next, using sine rule:
13/sin 59=c/sin 53.625
c=12.2113361
c~12.2111
From the second triangle:
a=13, b=14, A=49°, B=112.583° and C=180-(A+B)=8.417°
also;
a/sin A=c/sin C
13/sin 59=c/sin 8.417
c=2.219981
c~2.220
B=67.4,C=53.6, c=12.2,B=112.6, C=8.4,c=2.2
4] C=37°, a=19, c=8
a/sin A=c/sin C
19/sin A=8/sin 37
sin A/18=sin 37/8
sin A=1.354084
given that there is no sine inverse of 1.354084, then we conclude that there is no triangle formed.
5] B=36°, a=38, c=18
using cosine rule:
b²=a²+c²-2accos B
b²=38²+18²-2×38×18cos36
b²=661.2647517
b=25.715~25.7
using sine rule:
a/sin A=b/sin B
38/sin A=25.7/sin 36
sin A=0.8691
A=119.8
C=180-(A+B)
C=24.2°
the answer is b~25.7, C~24.2, A~119.8
6. A=46°, b=27 ft, c=14 ft
Area=1/2bcsin A
Area=1/2×27×14sin 46
Area=135.96 ft²
7] a=240, b=127, c=281
s=(a+b+c)/2
=(240+127+281)/2
=324
A=√s(s-a)(s-b)(s-c)
A=√324(324-240)(324-127)(324-281)
A=230,546,736
A=15,183.76554~15,183.77
8] When you plot thediagram, it can be seen that we have a right-angled triangle. Using the Pythagorean theorem we have:
d²=a²+b²
substituting the values we get:
d²=84²+135²
d²=25281
⇒d=√25281
d=159 km
9] P=(5,8) and Q=(6,9)
component is:
<6-5,9-8>
=<1,1>
magnitude will be:
║PQ║=√(1²+1²)=√2
the answer is <1,1>,√2
10] u=<-3,4>, v=<8,2>. The u+v will be:
u+v=<u1+v1, u2+v2>
where:
u=<u1,u2>
v=<v1,v2>
u+v=<-3+8,4+2>
=<5,6>
11] u=<3,-1>, v=<-6,-6>. Find 9u+2v.
9u=<9(3),9(-1)>
=<27,-9>
2u=<2(-6,-6)>
=<-12,-12>
then:
9u+2v=<27,-9>+<-12,-12>
=<(27-12),(-9-12)>
=<15,-19>
12. a=4i-4j, b=4i+5j. Then a*b will be:
Note that a*b is a scalar product.
given that:
a=xi1+yji; b=xi2+yj2
a*b=x1*x2+y1*y2
a*b=(4)*(4)+(-4)*(5)
=16-20
=-4
13] u=<-5,-4>, v=<-4,-3>. The angles between the vectors will be:
The angles will be calculated using the formula:
cos x=u*v/lullvl
u*v=<-5,-4>*<-4,-3>
=(-5)*(-4)+(-4)*(-3)
=20+12
=32
lul=l<-5,-4>l
=√(-5)²+(-4)²
=√41
lvl=l<-4,-3>l
=√(-4)²+(-3)²
=√25
=5
hence:
cos x=32/(5√41)
=0.9995~1
x=1.7899~1.8°
14] u=<6,-2>, v=<8,24>, the dot product will be:
(6×8)+(-2×24)
=48-48
=0
because the dot product is zero, then the vectors are said to be orthogonal.
15] r=<9,-7,-8>, v=<3,4,7>, w=<6,-9,7>
v*w
=(3*6)+(4*(-9))+(7*7)
=18-36+49
=31
16] Express -3i in trigonometric form
z=a+bi=lzl(cos θ+i sin θ)
lzl=sqrt(a^2+b^2)
a=0, b=-3
lzl=sqrt((-3)^2)
lzl=3
The angle of the point on a complex plane is normally the inverse tangent of the complex portion divide by the real portion.
θ=arctan(3/0)
because the value is undefined, the value of t=π/2
substituting the values ofθ=π/2 and lzl=3, we get:
3(cos(π/2))+isin(π/2)
=3(cos 90°+isin90°)
17] -3+3√3i is in the form of a+bi, where a=-3 and b=-3√3. The trigonometric form is
r(cosθ+isinθ)
r=√a²+b²
θ=arctan(b/a)
computing for r we get:
r=√(-3)²+(3√3)²
r=√9+27
r=√36
r=6
Next we findθ
θ=arctan(-3√3)/3=π/3=60° or 2/3π=120°
hence the trigonometric form will be:
6(cos 2π/3+isin 2π/3)
The answer is:
six times the quantity cosine of two pi divided by three plus i times sine of two pi divided by three
18. 5/[2(cos 150+isin 150)]
We know that:
150=180-30
This implies that 150° is in the 2nd quadrant and it makes an angle of 30° with the negative x-axis.
cos (150)=-(√3)/2
sin 150=1/2
hence our expression will be:
5/[2(-(√3)/2+i/2)]
=5/[√3+i]
rationalizing the denominator by multiplying the numerator and the denominator by√3-i we get:
[(√3-i)/(√3-1)]*[5/(√3+i)
this gives us
=[5*(√3-i)]/(3+1)
=5/4(√3-i)
multiplying through by -1 we get:
-5/4(√3)+-5/4i
the answer is:
negative five square root three divided by four plus five divided by four times i
19] To solve the question we assume that the angles are measured counter-clockwise from the positive x-axis.
The angle between the forces is 100-10=90
Based on this angle we can calculate the magnitude of the resultant force, the angles between the resultant forces and the 2 given forces. Thus, the magnitude of the resultant force using 25 and 30 we shall have:
√30²+25²
=√(900+625)
=√1525
=39.0513~39.05
The angle that resultant force R, makes with the 25 pound for will be:
tanθ=30/25=1.2
⇒θ~50.19°
the angle between the x-axis and resultant for will give us:
10+50.19
=60.19°
The magnitude of the resultant force is 39.05 pounds.
20. The cube root of 125(cos 288+isin288) will be:
∛[125(cos 288+i sin 288)]
=5∛[cos 288+isin 288]
from the rules of trigonmetry;
cos x+i sin x=e^i(x)
=5*∛(e^(i288))
=5*(e^(i*288))^(1/3)
=5*(e^(i288*1/3))
=5*e^(i96)
this can be re-written as:
5(cos 96+i sin 96)