Home Right Triangles Angles in Standard Position Sine Law & Ambiguous Case Cosine Law Problem Solving Summary&Test

UNIT 6  : BASIC TRIGONOMETRY WITH TRIANGLES

LESSON 1:  SOLVING RIGHT TRIANGLES HOMEWORK QUESTIONS  (Solutions below)

In each question below, round sides to nearest tenth and angles to nearest tenth of a degree.

1.  Solve the triangles in the diagrams below.

a)                                                                                                                                 b)

10.4

A                                                    C                                                                                                                  X

41o

c                              a

46 cm                             y

35o

B                                                                    Y                                                Z

x

2.  Solve the triangles in the diagrams below.

a)   < R = 90o                                                                                                               b)  < D = 90o

D

P

13.5 cm                                                                                                                                                       42.5 cm

Q                                                                                                                                              52.3 cm                                     F

11.6 cm

p

R

E

3.  Solve each of the given triangles.

4.  Find EF to the nearest tenth in the diagram below.

D

52.2o                                               30.4o

E                                                      G

F         12.2 cm

5.  Find < CAD in the diagram below.

A

15 cm

B

8 cm    C     11 cm           D

6.  Find < RTS in the diagram below.

P

T

12.4 cm

102.3o

Q          13.6 cm             R                                 S

7.  Find EF to the nearest tenth in the diagram below.

D

14.8 cm

57.1o                                               33.4o

E                                                      G

F

Solutions:

1.  Solve the triangles in the diagrams below.

a)                                                                                                                                 b)

10.4

A                                                    C                                                                                                                  X

41o

c                              a

46 cm                             y

35o

B                                                                    Y                                                Z

x

Solution 1(a):  Find side “a” first.

a = side opposite  and  10.4 = side adjacent relative to 41o.  Hence we use the tangent ratio since it involves opposite and adjacent.

Now find side “c”:

c = hypotenuse and  10.4 = side adjacent relative to 41o.  Hence we use the secant ratio since it involves hypotenuse and adjacent.

To find < B, we note that < C + < B = 90o and hence < B = 90 – 41 = 49o

Solution 1(b):  Find side “y” first.

y = side opposite  and  46 = hypotenuse relative to 35o.  Hence we use the sine ratio since it involves opposite and hypotenuse.

Now find side “x”:

46 = hypotenuse and  x = side adjacent relative to 35o.  Hence we use the cosine ratio since it involves hypotenuse and adjacent.

To find < X, we note that < X + < Y = 90o and hence < X = 90 – 35 = 55o

2.  Solve the triangles in the diagrams below.

a)   < R = 90o                                                                                                               b)  < D = 90o

D

P

13.5 cm                                                                                                                                                       42.5 cm

Q                                                                                                                                              52.3 cm                                     F

11.6 cm

p

R                                                                                                                           E

Solution 2(a):  Find <P first.

13.5 = hypotenuse  and  11.6 = side adjacent relative to <P.  Hence we use the cosine ratio since it involves hypotenuse and adjacent.

To find < Q, we note that < P + < Q = 90o and hence < Q = 90 – 30.8 = 59.2o

Now find side “p”:

13.5 = hypotenuse and  p = side opposite relative to 30.8o.  Hence we use the sine ratio since it involves hypotenuse and opposite.

Solution 2(b):  Find <F first.

52.3 = opposite  and  42.5 = side adjacent relative to <F.  Hence we use the tangent ratio since it involves opposite and adjacent.

To find < E, we note that < E + < F = 90o and hence < E = 90 – 50.9 = 39.1o

Now find side “d”:

d = hypotenuse and  52.3 = side opposite relative to 50.9o.  Hence we use the cosecant ratio since it involves hypotenuse and opposite.

3.  Solve each of the given triangles.

14.2

A                                                    C

18.8                              a

B

Solution 3(a):  Since we have 2 sides given, we find <A first.

18.8 = hypotenuse  and  14.2 = side adjacent relative to <A.  Hence we use the cosine ratio since it involves hypotenuse and adjacent.

To find < B, we note that < A + < B = 90o and hence < B = 90 – 40.9 = 49.1o

Now find side “a”:

18.8 = hypotenuse and  a = side opposite relative to <A = 40.9o.  Hence we use the sine ratio since it involves hypotenuse and opposite.

f

D                                                    E

e                               15.4

24.2o

F

Solution 3(b):  Since we are given a side and angle, we find a side first. Find side “e”.

15.4 = side adjacent and  e = hypotenuse relative to 24.2o.  Hence we use the secant ratio since it involves adjacent and hypotenuse.

Now find side “f”:

f = side opposite and  15.4 = side adjacent relative to 24.2o.  Hence we use the tangent ratio since it involves opposite and adjacent.

To find < D, we note that < D + < F = 90o and hence < D = 90 – 24.2 = 65.8o

16.4

P                                                     M

26.8o

m                               p

N

Solution 3(c):  Since we are given a side and angle, we find a side first.

16.4 = side adjacent and  m = hypotenuse relative to 26.8o.  Hence we use the secant ratio since it involves adjacent and hypotenuse.

Now find side “p”:

p = side opposite and  16.4 = side adjacent relative to 26.8o.  Hence we use the tangent ratio since it involves opposite and adjacent.

To find < N, we note that < N + < P = 90o and hence < N = 90 – 26.8 = 63.2o

22.4

A                                                    B

b                                33.6

C

Solution 3(d):  Since we have 2 sides given, we find <A first.

33.6 = side opposite  and  22.4 = side adjacent relative to <A.  Hence we use the tangent ratio since it involves opposite  and adjacent.

To find < B, we note that < A + < B = 90o and hence < B = 90 – 56.3 = 33.7o

Now find side “b”:

b = hypotenuse and  33.6 = side opposite relative to <A = 56.3o.  Hence we use the cosecant ratio since it involves hypotenuse and opposite.

4.  Find EF to the nearest tenth in the diagram below.

D

52.2o                                               30.4o

E                                                      G

F         12.2 cm

Solution:

We must use the triangle with 3 elements known.  Hence we first find side DF in this triangle.

5.  Find < CAD in the diagram below.

A

15 cm

B

8 cm    C     11 cm           D

Solution:

6.  Find < RTS in the diagram below.

P

T

12.4 cm

102.3o

Q          13.6 cm             R                                 S

Solution:

Again we must use the triangle with 3 elements given.

7.  Find EF to the nearest tenth in the diagram below.

D

14.8 cm

57.1o                                               33.4o

E                                                      G

F

Solution:

We must use the triangle with 3 elements known.  Hence we first find side DF in this triangle.