Question #225050

1. Two lighthouses A and B are 25km apart and A is due west of B. A submarine S is on a bearing of 137degrees from A and on a bearing of 170degrees from B. Find the distance of S from A and the distance of S from B.


2. In the triangle PQR, PQ= 7cm, PR= 8cm and QR= 9cm. Find angle QPR.\


3. Find the smallest angle in a triangle whose sides are of length 3x, 4x and 6x.


4. Shop Y is 9km due North of shop Z. Shop X is 8km from Y, 5km from Z and somewhere to the west of line YZ.

a) Draw the triangle XYZ and find angle YZX.

b) To improve sales, shop X is moved due to South until it is due West of Z. Find how far it moved.


1
Expert's answer
2021-08-11T13:29:22-0400

1.


Using Sine rule:


asin(A)=bsin(B)\frac{a}{sin\left(A\right)}=\frac{b}{sin\left(B\right)}


25sin33=bsin47\frac{25}{sin\:33}=\frac{b}{sin\:47}


b=25×sin47sin33b=\frac{25\times sin\:47}{sin\:33}


b=33.57kmb=33.57km


Hence, distance of S to B is 33.57km.


bsin(B)=csin(C)\frac{b}{sin\left(B\right)}=\frac{c}{sin\left(C\right)}


33.57sin(47)=csin(100)\frac{33.57}{sin\left(47\right)}=\frac{c}{sin\left(100\right)}


c=33.57×sin100sin47c=\frac{33.57\times sin\:100}{sin\:47}


c=45.20kmc=45.20km


Hence, the distance of S to A is 45.20km.


2.


To find angle QPR we use the cosine rule:

a2=b2+c22bcCosAa^2\:=b^2+c^2-2bcCosA


CosA=b2+c2a22bcCos\:A=\frac{b^2+c^2-a^2}{2bc}


CosA=72+82922×7×8Cos\:A=\frac{7^2+8^2-9^2}{2\times 7\times 8}


CosA=0.2857Cos\:A=0.2857


A=Cos1(0.2857)A=Cos^{-1}\left(0.2857\right)

A73.4oA\approx 73.4^o


Hence angle QPR is 73.4o73.4^o


3.

Angle1=3xAngle\:1=3x

Angle2=4xAngle\:2=4x

Angle3=6xAngle\:3=6x


Anglesumpropertyofatriangle=180Angle\:sum\:property\:of\:a\:triangle=180

3x+6x+4x=180\:\:3x+6x+4x=180

13x=18013x=180

x=13.85x=13.85


The smallest angle will be: 3x3x

3x=3×13.8541.5o3x=3\times13.85\approx41.5^o


4.


(a). The angle YZX can be determined by cosine rule:

a2=b2+c22bcCosAa^2\:=b^2+c^2-2bcCosA


CosA=b2+c2a22bcCos\:A=\frac{b^2+c^2-a^2}{2bc}


CosA=52+92822×5×9Cos\:A=\frac{5^2+9^2-8^2}{2\times 5\times 9}


A=Cos1(0.4667)A=Cos^{-1}\left(0.4667\right)


A62.2oA\approx62.2^o


Hence, angle YZX is 62.2o62.2^o


(b).


To determine the distance Shop X moved to a reach a new location we use the formula:

Distance=θ360×2×π×rDistance=\frac{\theta }{360}\:\times 2\times \pi \times r\:


Distance=207.8o360o×2×227×5Distance=\frac{207.8^o}{360^o}\:\times \:2\times \:\frac{22}{7}\:\times \:5


Distance18.14kmDistance\approx 18.14km


Hence, shop X moved 18.14 km to reach a new location.

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