McGraw Hill Pre Calculus 11 Student Edition Chapter 2 Exercise 2.4 Trigonometry

Pre-Calculus 11 Student Edition Chapter 2 Trigonometry

Page 114 Problem 1 Answer

The objective of the problem is to draw △ABC

with a given length of sides:

a=3 cm

b=4 cm

c=5 cm

​The required triangle with given side lengths can be done with a ruler as shown below:

Hence, the required triangle with given side values can be shown below:

Page 114 Problem 2 Answer

The values of a²,b², and c².

The given values of a=3,b=4, and c=5

The value of

a²=(3)²=9

b²=(4)²=16

c²=(5)²=25​

The values are :

a²=9,b²=16,c²=25

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Page 114 Problem 3 Answer

On comparing the values of a²,b²,c².

The values are :a²=9,b²=16,c²=25.

The value of −a²=−9

The relation is true when −9+16<25.

Hence, the relation is true when a2+b2 <c².

The relation is true when −a²+b²<c².

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Page 114 Problem 4 Answer

The measure of ∠C.

Using Pythagorean Theorem, a²+b²=c²

Here, the given values are a=3,b=4,c=5

This triangle follows the Pythagorean triplet rule,

c²=a²+b²

5²=3²+4²

It is a right-angled triangle at C.

So, ∠C=90˚

The value of ∠C=90˚.

Page 114 Problem 5 Answer

Draw an acute ΔABC, Acute angle is less than 90∘

So we make a triangle with all the angles less than 90∘

For example , we can make a triangle with angle ∠A=56∘and angle∠B=74∘and angle ∠C=50∘

All angles are less than 90∘.

Acute triangle ABC

Page 114 Problem 6 Answer

From part (a) we got the acute triangle ABC

Make the base that is length of a= 5 cm. Using protractor measure the angle and connect all the vertices.

Now measure the length of b  and c

Length of b=5.8  and c=4.6

Length of sides of triangle ABC

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a=5 cm

b= 5.8 cm

c=4.6 cm

Page 114 Problem 7 Answer

Given: a=5,b=5.8,c=4.6

To find: a²,b²,c²

a²=5×5=25

b²=5.8×5.8=33.64

c²=4.6×4.6=21.16

a²=25,b²=33.64,c²=21.16

Page 114 Problem 8 Answer

Given: a²=25,b²=33.64,c²=21.26

To compare: a²+b²and c²

a²+b²=25+33.64=58.64

So: a²+b²>c² is true.

We do not get the other statement true after calculations.

a²+b²>c² is true.

Page 115 Problem 9 Answer

Given: a²+b²=25,c²=25

To find: 2abcosC=a²+b²−c²

2abcosC=25-25=0

2ab cos C = 0

Page 115 Problem 10 Answer

Given: a²+b²=58.64,c²=21.26

To find: 2abcosC

2ab cos C = 58.64 − 21.26=37.38

2ab cos C = 37.38

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Page 115 Problem 11 Answer

We have given a triangle ΔABC:-

We have to assume different values of a,b,c and complete the following table :-

Firstly we have given that a=3,b=4,c=5.

then we have :-

c²=5²⇒c²=25 and

a²+b²=3²+4²

⇒a2+b2=9+16

⇒a²+b²=25

Now using the cosine laws, we have :-

c²=a²+b²−2abcosC

⇒2abcosC=a²+b²−c²

Then by putting the values we have :-

2abcosC=25−25⇒2abcosC=0

​Now assume that :-

a=4,b=5,c=6

Then we have :-

c²=6²⇒c²=36 and

a²+b²=4²+5²

⇒a²+b²=16+25

⇒a²+b²=41 and

2abcosC=a²+b²−c²

⇒2abcosC=41−36

⇒2abcosC=5

​Now assume that:-

a=5,b=6,c=7, then we have :-

c²=7²⇒c²=49 and

a²+b²=5²+6²

⇒a²+b²=25+36

⇒a²+b²=61 and

2abcosC=a²+b²−c²

⇒2abcosC=61−49

⇒2abcosC=12

​Now assume that :-a=5,b=6,c=6

Then we have:-c²=62⇒c²=36 and

a²+b²=5²+6²⇒a²+b²

=25+36⇒a²+b²=61 and

2abcosC=a²+b²−c²

⇒2abcosC=61−36

⇒2abcosC=25​

Complete the table :-

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The required completed table is as shown below :-

Page 115 Problem 12 Answer

Draw a triangle ABC  with angle C obtuse.

Let us draw a triangle with angle C= 94 degrees

The sides are a=3, b=5  and c=6

The equation we got from step 4 is

c²=a²+b²−2abcos(C)

Substitute the values

6²=3²+5²−2(3)(5)cos(94)

36=9+25−(−2)

36=36 True

​The equation we got in step 4 hold true for the triangle with angle C= 95, a=3, b=5  and c=6

Page 117 Problem 13 Answer

Given: The distance from point C to A is 35.5 m and from C to B is 48.8 m.

Also, the angle at C is 54∘.

To find: The distance AB.

Use cosine law to find the required distance.

First, draw the diagram for better understanding.

Use cosine law a²=b²+c²−2bccosA to find the distance AB. Herea=AB,b=35.5,c=48.8 and A=54∘.

Substitute all the values in the formula and simplify to find the distance AB.

a²=b²+c²−2bccosA

AB²=35.52+48.82−2(35.5)(48.8)cos54∘

AB²≈1260.25+2381.44−3464.8(0.58778)

AB²≈1260.25+2381.44−2036.5583

AB²≈1605.1316

AB≈√1605.1316

AB≈40.1

​Hence the distance AB is approximately 40.1 m.

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Page 119 Problem 14 Answer

Given: In triangle ABC ,a=9,b=7 and ∠C=33.6∘.

To find: The length of the unknown side and the measures of the unknown angles.

Use cosine law to determine the length of the unknown side and the measures of the unknown angles.

First, sketch the diagram of the triangle ABC.

Now find the length of the side c by using cosine law. Substitute a=9,b=7 and ∠C=33.6∘in the cosine law and simplify.

c²=a²+b²−2abcosC

c²=9²+7²−2(9)(7)cos33.6∘

c²=81+49−126(0.832…)

c²=81+49−104.948…

c²=25.051…

c=√25.051…

c=5

​Now find the unknown angles of the triangle.

Substitute a=9,b=7 and c=5 in the cosine law and simplify.

cosB=a²+c²−b²

2ac/cosB=9²+5²−7^2/2(9)(5)

cosB=81+25−49/90

cosB=57/90

∠B=cos−1(57/90)

∠B=cos−1/(0.633…)

∠B=50.703∘…

Thus the measure of∠B is approximately 50.7∘.

Use the angle sum property of a triangle to find the measure of angle A

∠A+∠B+∠C=180∘

∠A+50.7∘+33.6∘=180∘

∠A=180∘−(50.7∘+33.6∘)

∠A=180∘−84.3∘

∠A=95.7∘

Thus the measure of∠A is 95.7∘.

Hence the length of side c is 5 and the measures of unknown angles are∠A=95.7∘ and ∠B=50.7∘.

Page 119 Problem 15 Answer

The given figure is:

We need to find the length of the third side of the given triangle.

We will use the Law of Cosine to find the length of the third side of the given triangle.

In the given triangle, we have a=14,b=9 and C=17∘.

Using the Law of Cosine, we get

c²=(14)²+(9)²−2(14)(9)cos(17∘)

c²=196+81−252(0.9563)

c²=277−240.9876

c²=36.0124

c=√36.0124

c≈6

The length of the third side of the given triangle is about 6 cm.

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Page 119 Problem 16 Answer

The given figure is:

We need to find the length of the third side of the given triangle, i.e., MN.

We will use the Law of Cosine to find the length of the third side of the given triangle.

In triangle MNL,NL=13,ML=29 and ∠L=41∘. So, we have a=13,b=29 and C=41∘.

Using the Law of Cosine, we get

c²=(13)²+(29)²−2(13)(29)cos(41∘)

c²=169+841−754(0.7547)

c²=1010−569.0438

c²=440.9562

c=√440.9562

c≈21

The length of the third side of the given triangle is about 21 mm.

Page 119 Problem 17 Answer

The given figure is:

We need to find the length of the third side of the given triangle, i.e., DE.

We will use the Law of Cosine to find the length of the third side of the given triangle.

In the given triangle DEF,EF=21,DF=30,m∠F=123∘..So, we have a=21,b=30 and C=123∘.

Using the Law of Cosine, we get

c²=(21)²+(30)²−2(21)(30)cos(123∘)

c²=441+900−1260(−0.544639)

c²=1341+686.24514

c²=2027.24514

c=√2027.24514

c≈45.02

The length of the third side of the given triangle is about 45.02 m.

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Page 119 Problem 18 Answer

The given figure is:

We need to find the measure of ∠J.

We will use the Law of Cosine to find the measure of the required angle.

In the given triangle the opposite side of angle J is IH=10 m. So, we have a=10,b=11,c=17.

Using the Law of Cosine, we get

cosJ=(11)²+(17)²−(10)²/2(11)(17)

cosJ=121+289−100/374

cosJ=310/374

J=cos−1(310/374)

J≈34.02

The measure of angle J is 34.02∘.

Page 119 Problem 19 Answer

The given figure is:

We need to find the measure of the angle L.

We will use the Law of Cosine to find the measure of the required angle.

In the given triangle the opposite side of angle L is MN=18 cm. So, we have a=18, b=10.4, c=21.9.

Using the Law of Cosine, we get

cosL=(10.4)²+(21.9)²−(18)²/2(10.4)(21.9)

cosL=108.16+479.61−324/455.52

cosL=263.77/455.52

L=cos−1/(263.77/455.52)

L≈54.62∘

The measure of angle L is 54.62∘.

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Page 119 Problem 20 Answer

The given figure is:

We need to find the measure of angle P.

We will use the Law of Cosine to find the measure of the required angle.

In the given triangle the opposite side of angle P is QR=14 mm. So, we have a=14,b=6,c=9.

Using the Law of Cosine, we get

cosP=(6)²+(9)²−(14)²/2(6)(9)

cosP=36+81−196/108

cosP=−79/108

P=cos−1(−79/108)

P≈137.01∘

The measure of angle P is 137.01∘.

Page 119 Problem 21 Answer

The given figure is:

We need to find the measure of angle C.

We will use the Law of Cosine to find the measure of the required angle.

In the given triangle the opposite side of the angle C is AB=31 m. So, we have a=20,b=13,c=31.

Using the Law of Cosine, we get

cosC=(20)²+(13)²−(31)²/2(20)(13)

cosC=400+169−961/520

cosC=−392/520

C=cos−1(−392/520)

C≈138.92∘

The measure of angle C is 138.92∘.

Page 120 Problem 22 Answer

Given: In a triangle PQR,PQ=29,PR=28 and ∠P=52∘.

To find: The length of the unknown side and the measures of the unknown angles.

Use cosine law to determine the length of the unknown side and the measures of the unknown angles.

For finding the unknown side, we will use the cosine law.

Applying cosine law, the length of p will be,

p²=q²+r²−2qrcosP

⇒p²=28²+29²−2×28×29×cos520

⇒p²=784+841−1624×0.6157

⇒p²=1625−999.8968

⇒p²=625.1032

⇒p=25.00206

Hence, p=25km (approx)

We will find ∠Q by using formula

cosQ=p²+r²−q²/2pr

⇒cosQ=25²+29²−28²/2×25×29

⇒cosQ=625+841−784/1450

⇒cosQ=682/1450

⇒cosQ=0.470345

⇒Q=cos−1/(0.470345)

⇒∠Q=620

Now, ∠R=1800−∠Q−∠P

∠R=1800−620−520

∠R=660

Hence, The missing angles are 620 and 520

The missing angles are 620 and 520.

The missing side is 25km

Page 120 Problem 23 Answer

Given that : In ΔRST,

Sidesr=5cm,s=9.1cm,t=6.8cm

To Find :∠R,∠S,∠T

Strategy: We will use cosine law to find the angles of triangle.

or ∠R, we have

cosR=s²+t²−r²/2st

⇒cosR=9.12+6.82−52/2×9.1×6.8

⇒cosR=82.81+46.24−25/123.76

⇒cosR=104.05/123.76

⇒cosR=0.84074

⇒R=cos−1/(0.84074)

 

⇒R=32.78  (approx)                  Using cosine inverse table

⇒∠R=330

For ∠S, we have

cosS=r²+t²−s²/2rt

⇒cosS=5²+6.8²−9.1²/2×5×6.8

⇒cosS=25+46.24−82.81/2×5×6.8

⇒cosS=−11.57/68

⇒cosS=−0.17015

⇒S=cos−1(−0.170150)

⇒S=99.796 (approx)                    using cosine inverse table

Hence, ∠S=1000

We will use the following relation to find ∠T:∠R+∠S+∠T=1800

⇒330+1000+∠T=1800

⇒1330+∠T=1800

⇒∠T=1800−1330

⇒∠T=470

Hence, ∠T=470

he measure of angles of ΔRST is∠R=330,∠S=1000,∠T=470.

Page 120 Problem 24 Answer

Given: In a triangle ABC, AB=24, AC=34 and ∠A=67∘.

To find: The length of BC.

Use cosine law to determine the required length.

First sketch the triangle ABC.

SubstituteAB=24, AC=34 and ∠A=67∘in the cosine law and simplify.

BC²=AB²+AC²−2(AB)(AC)cosA

BC²=24²+34²−2(24)(34)cos67∘

BC²=576+1156−1632(0.390…)

BC²=576+1156−637.673…

BC²=1094.326…

BC=√1094.326…

BC=33

​Hence the length of BC is approximately33 m.

Page 120 Problem 25 Answer

Given: In a triangle ABC, AB=15,BC=8 and ∠B=24∘.

To find: The length ofAC.

Use cosine law to determine the required length.

First sketch the triangle ABC.

SubstituteAB=15,BC=8 and ∠B=24∘in the cosine law and simplify to find the length of AC.

AC²=AB²+BC²−2(AB)(BC)cosB

AC²=15²+8²−2(15)(8)cos24∘

AC²=225+64−240(0.913…)

AC²=225+64−219.250…

AC²=69.749…

AC=√69.749…

AC=8.4

​Hence the length of AC is approximately 8.4 m.

Page 120 Problem 26 Answer

Given: AC  = 10 cm, BC = 9 cm, and ∠C = 48°. To Determine the length of AB.By using cosine law

We have triangle ABC with  AC  = 10 cm, BC = 9 cm, and ∠C = 48°

So according to the cosine law

​AB²=10²+9²−2(10)(9)cos48∘

AB²=60.556…

AB²=√60.556…

AB=7.781…

AB=7.8 cm , to the nearest tenth of a centimetre

​Hence, the length of AB. is 7.8 cm

Page 120 Problem 27 Answer

Given:  AB  = 9 m, AC = 12 m, and BC = 15 m.To  Determine the measure of ∠B.By using sine ratio

We have AB  = 9 m, AC = 12 m, and BC = 15 m.

△ABC is a right triangle, because 15²=9²+12²

Use the sine ratio.

sin ∠B∠B

=12/15

=sin−1(12/15)

=53.130…

R=53∘

∠B=53∘, to the nearest degree.

Hence, the measure of ∠B is 53o

Page 120 Problem 28 Answer

Given: AB  = 18.4 m, BC = 9.6 m, and AC = 10.8 m. To Determine the measure of ∠A.By using cosine law

We have AB  = 18.4 m, BC = 9.6 m, and AC = 10.8 m.

using the cosine law

​​

9.62=10.82+18.42−2(10.8)(18.4)cosA

92.19=116.64+338.56−397.44cosA

397.44cosA=455.2−92.19

cosA=363.01/397.44

∠A=cos−1(363.01/397.44)

∠A=24.024…​

∠A=24∘, to the nearest degree.

Hence, the measure of ∠A  is 24o

Page 120 Problem 29 Answer

Given: AB  = 4.6 m, BC = 3.2 m, and AC = 2.5 m. Determine the measure of ∠C.

To Determine the measure of ∠C.By using cosine law

We have AB  = 4.6 m, BC = 3.2 m, and AC = 2.5 m.

Using cosine law

4.62=3.22+2.52−2(3.2)(2.5)cosC

21.16=10.24+6.25−16cosC

16cosC=16.49−21.16

cosC=−4.67/16

∠C=cos−1(−4.67/16)

∠C=106.970…

∠C=107∘, to the nearest degree.

​Hence, measure of ∠C is  ∠C=107,