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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

41^o

c a

46 cm y

35^o

B Y Z

x

2. Solve the triangles in the diagrams below.

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

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

52.2^o 30.4^o

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

102.3^o

Q 13.6 cm R S

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

D

14.8 cm

57.1^o 33.4^o

E G

F

Solutions:

1. Solve the triangles in the diagrams below.

a) b)

10.4

A C X

41^o

c a

46 cm y

35^o

B Y Z

x

Solution 1(a): Find side “a” first.

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

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

Solution 1(b): Find side “y” first.

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

Now find side “x”:

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

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

2. Solve the triangles in the diagrams below.

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

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 = 90^o and hence < Q = 90 – 30.8 = 59.2^o

Now find side “p”:

13.5 = hypotenuse and p = side opposite relative to 30.8^o. 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 = 90^o and hence < E = 90 – 50.9 = 39.1^o

Now find side “d”:

d = hypotenuse and 52.3 = side opposite relative to 50.9^o. 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 = 90^o and hence < B = 90 – 40.9 = 49.1^o

Now find side “a”:

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

f

D E

e 15.4

24.2^o

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 24.2^o. 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 24.2^o. Hence we use the tangent ratio since it involves opposite and adjacent.

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

16.4

P M

26.8^o

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 26.8^o. 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 26.8^o. Hence we use the tangent ratio since it involves opposite and adjacent.

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

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 = 90^o and hence < B = 90 – 56.3 = 33.7^o

Now find side “b”:

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

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

D

52.2^o 30.4^o

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

102.3^o

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

57.1^o 33.4^o

E G

F

Solution:

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

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