How to Solve These Basic Trigonometry Questions (Bearings, Word Problems)?

Hi All,
I know you are all busy, but I don't understand how to solve these, please help? Please fully answer all parts of the question, thank you!

Question #5
enter image source here

Question #18 and #19
enter image source here

Question #3 and ...

Hi All,
I know you are all busy, but I don't understand how to solve these, please help? Please fully answer all parts of the question, thank you!

Question #5
enter image source here

Question #18 and #19
enter image source here

Question #3 and #4
enter image source here

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1
Feb 17, 2017

Answer:

5.#=7km 810 m 25cm#, nearly. 18. #cos^(-1)(3/4)=41.41^o#, nearly.

19, #53.18^o#

Explanation:

  1. Let A and B start from O.

After half an hour,

#OA = (time xx speed) =1/2(10)=5 km#. Likewise,

#OB = 1/2(12)=6 km#, and so,

#AB=sqrt((OA)^2+(OB)^2)=sqrt(5^2+6^2)=sqrt61=7.81025 km#

#=7km 810 m 25cm#, nearly.

  1. Let the base be BC = b and equal sides AB = AC = a.

Then the equal sides length #a = 2/3b#

#cos (base angle) = cos B =(b/2)/a=3/4.

So, #angleB=cos^(-1)(3/4)=41.41^o#, nearly.

  1. Let the altitude be AN and base BC.

In #triangle BNA# that is right angled at N,

tan B = AN/BN=AN/14.

So, the altitude AN = 14 tan B=14 tan 24^o#.

#Area = 1/2(base)xx#(altitude)#=1/2(28)(14tan 24^o)=196 tan24^o#.

Treble this area =588tan24^o.

For this area,

the corresponding same-base base angle B is given by

#Area = 1/2(28)(14)tanB=588tan24^o#

So, #tan B = 3tan24^o=1.3357 and B =53.18^o#, nearly.

For this approximation, I have used calculator

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dk_ch Share
Feb 21, 2017

Question 5

drawn

In the above figure O is the starting point. A and B are the positions of two runners after 30 min or 0.5hour running @ 10km/h towards north and @12km/h towards east respectively.

So #OA=10xx0.5=5km and OB=12xx0.5=6km#

By Pythagorean theorem

The distance of runner B from A

#AB =sqrt(OA^2+OB^2)=sqrt(5^2+6^2)=sqrt61km#

Bearing is always measured in clockwise direction w.r. to north line (shown in figure by red arrow)

So the bearing of B from A

#=180^@-tan/_BAO=180^@-tan^-1(6/5)=(180-50)^@=130^@#

Question 18

drawn

The given triangle is isosceles in which the equal sides are #2/3# of the base. So let us consider an isosceles triangle ABC where the base BC is 6 unit and the equal sides are 4 unit . The base angle is # theta#. AD is perpendicular from A to BC.

It is obvious from figure that #costheta = "adjacent"/"hypotenuse" =3/4#

So #theta = cos^-1(3/4)=41.4^@#

Question 19

As per given condition of the question the second isosceles triangle (EBC) has same base that of first one (ABC) but the area of second one is thrice that of first one. It is possible only if the height of second triangle is thrice that of first one. Since the area of the triangle is proportional to height when base is constant.

This has been shown in the fig below.

The perpendicular drawn from vertex of an isosceles triangle bisects the base.

drawn

From fig

#(DeltaEBC)/(DeltaABC)=(1/2xxBCxxED)/(1/2xxBCxxAD) #

#=>3=(ED)/(AD)#

#ED=3AD#

Now #(tan/_ECB)/(tan/_ACB)=((ED)/(BC))/((AD)/(BC))=(ED)/(AD)#

#=>(tantheta/tan24^@ )=3#

#=>tantheta=3xxtan24^@=1.34#

#=>theta =tan^-1(1.34)~~53.2^@#

Question 3a

drawn

Bearing

I) #B" from "A->41^@#

II) #C" from "B->142^@#

III) #B" from "C->(279+43)^@=322^@#

IV) #C" from "A->(41+58)^@=99^@#

V) #A" from "B->(142+38+41)^@=221^@#

VI) #A" from "C->279^@#

Question 3b

enter image source here

Bearing

I) #B" from "A->27^@#

II) #C" from "B->151^@#

III) #B" from "C->(246+85)^@=331^@#

IV) #C" from "A->(27+39)^@=66^@#

V) #A" from "B->(151+29+27)^@=207^@#

VI) #A" from "C->246^@#

Question 4

drawn

Bearing =# 90^@ +tan^-1(9/14)~~90^@+33^@=123^@#

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