LESSON 3.

0 SOLUTIONS OF RIGHT TRIANGLES

3.1 SOLVING RIGHT TRIANGLES
Six Parts of a triangle:
1. Three sides
2. three angles
SOLVING THE RIGHT TRIANGLES
Figure 5 shows a right triangle with its right angle at C. A and B are the acute angles; a
and b are the respective opposite sides.
To solve for the unknown parts, apply the following:
i.)

Trigonometric function (parts involved: Angle, 2 sides)

ii.)

Pythagorean theorem (parts involved: 3 sides)

B
c

a
C

Example 9: Solve the right triangle ABC in which a=12 and c=13
B

13

A
Solution:

12

a. Solve for b by Pythagorean theorem:

b=
=

=
=
b=5
b. Solve for A by trigonometric function
tan A =

A = tan-1

c. Solve for B.

= 672248"

Apply: A+B = 90

B = 90- A
= 90 - 67248"
B = 223712"

Check: A + B + C =180
3510

672248 + 22 3712 + 90 = 180

180 = 180

Example 10: Solve the right triangle ABC in which B = 3510 & c = 72.5
B
Solution:
72.5
A

a) Solve for A:

Apply:

A + B = 90
A = 90-B
= 90-3510
A = 5450

b) Solve for b: By trigonometric function

sin B =
b = c sinB

= 72.5 sin 3510

b = 41.8
c) Solve for a: By trigonometric function
cos B =
a = c cos B
= 72.5 cos 3510
a = 59.3
Check:
tan B =
tan 3510 =

0.705

= 0.705

Example 11: The base of an isosceles triangle is 20.6 and the base angles are 4659.
Find the equal sides and the height of the triangle.
Solution: Consider a triangle ABD
B

46059
A

C
10.30

Let :

BD bisects AC
i)

= cos A

AB =

10.30

= 15.10

i)

= tan A

BD

= 10.3 tan 4659

BD

= 11.04

The equal sides are each 15.10 and the altitude is 11.04 of the triangle.
Example 12: Find the perimeter and area of a regular nonagon inscribe I a circle of
radius18ft.

20
O

Let

20
9 ft

OA OB
AOB =

If

OC

40

bisects , AOB, then 2 right triangles are formed.

i)

Solve for each side of the nonagon:

Take the right triangle BOC: sin 20 =

BC = 9 sin 20
BC = 3.078 ft = AC
Hence the perimeter = 18 (3.078) = 55.404 ft
ii)

Solve for the area of one triangle: Take the triangle AOB

40

6.156

Since

OC = 100 cos 20

Thus, Area of triangle AOB =

(AB) (OC)

(6.156) (50 cos20)

= 289.237 sq. ft.

Area of nonagon = 9 (289,237) = 2,603.133 ft.
Example 13 : A long piece of sheet metal 1.3 meters wide is to be made into a trough of
trapezoidal cross-section with base 0.50 meter wide (see the figure
below).
h

a) Express the area of trapezoidal cross-section as a function of .

b) If = 5814, find the area of the cross-section.

Solution :
a) Since the base is 0.5 m, each the inclined sides has a length of 0.6 m
AB = 2x + 0.5
x = 0.6 cos
h = 0.6 sin
Hence, AB = 2(0.6 cos ) + 0.5
Recall:
Area of trapezoid =

(b1+b2)h

[0.5 +2(0.6 cos )+ 0.5](0.6 sin )

=
=

(1+1.2 cos ) (0.6 sin )

= 0.6 sin (0.5 + 0.6 cos )

b) if = 5814, area = 0.6 sin 5814 (0.5 + 0.6 cos 5814)
60

Area = 0.146 sq. meters

EXAMPLE 14:

A ladder leans against the side of a building with its foot 15 ft. from the
building. How far from the ground is the top of the ladder and how long
is the ladder if it makes an angle of 60 with the ground?

Solution:
B

C
15 ft

Let: BC = be the distance of the top of the ladder from the ground.
AB = be the length of the ladder.
To solve for BC: By trigonometric function
tan A =

tan 60 =

BC = 15 tan60
BC = 25.98 ft.
To solve for AB:
cos A =

AB =

Example 5:

= 30 ft.
A ladder leans against a vertical wall with its foot 15 feet from the wall. If
the ladder makes an angle of 60 with the ground,
a. how long is the ladder?
b. how far above the ground is the top of the ladder?

Solution:
Let:
a.

c = be the length of the ladder

a = be the height of its top from the ground
cos 60 =

60

15 ft

25

c =

c = 30 ft.
a
15
a = 15 tan 60
a = 15
or 25.98 ft.

b.

tan 60 =

Example 6: When the sun is 25 above the horizon, how long is the shadow cast by a
tree 150 ft. high?
Solution :

ft

A
Let:

A = 25
BC = 150 ft.
AC = be the length of the shadow cast by a tree

BC 150
Thus, tan 25 = AC = AC
150
AC = tan 25

AC = 321.68 ft

150

Exercise 2C
Lesson 2.6: Solution of the Right Triangle

Name: __________________________________ Score: ____________

Course/Section: ___________________________ Date: _____________
Solve for the right triangle ABC in which c = 90.
1. a = 2,

b=3

2. a = 643,

c = 751

3. b = 11,

c = 15.3

4. b = 12.4,

A = 15

5. b = 6.3,

B = 7825

6. c = 38,

B = 0.7

7. a = 6.7,

B = 43.8

8. c = 3.8,

A = 1.12

9. a = 38,

A = 0.6

10. a = 40,

B = 35.2

Evaluate in both degrees and radians. Use calculator only when necessary.
11. sin 1

12. cos -1 (0.1)

13. tan -1 (100)

14. arcsin (0.9)

15. arc cos

16. arc tan (0.8)

17. sec -1

18. arc csc

19. cot-1 (-

20. tan (tan-1 0.6)

21. If a = 11 and c =15.3, find sin and then find

(in degrees)

22. If b = 37.2 and c = 129.48, find cos and then find (in radians)

23. If a = 24 and b = 143, find tan and then find

(in degrees)

24. If a = 0.264 and c = 0.265, find sin , and then find (in radians)

25. If b = 9.5 and c = 19.3, find cos and then find (in degrees)

26. If a = 15 and b = 34.2, find tan and then find (in radian)

27. If a vertical post subtends an angle of 13 15 from a point on the ground 76 ft.
away, how high is the post?

3.2 ANGLE OF ELEVATION AND DEPRESSION

Definition:
ANGLE OF ELEVATION (Angle HOE)
= the vertical angle measured counterclockwise from the horizontal line
(Figure 5)
ANGLE OF DEPRESSION (Angle HOD)
= the vertical angle measured clockwise from the horizontal line
(Figure 5)

Fig. 5

Remarks: The word elevation indicates that E is higher than O. And depression
indicates that D is below O.
EXAMPLE 15: From the top of a tower at C, the angle of depression of point A on the
ground is 52. If distance AB is 165 meters, how high is the tower?
Solution:
Let BC = be the height of the tower

BC = 165 tan 52
BC = 211.19 m or 211m

Sun

BC = ?

EXAMPLE 16: A tree 150 ft tall casts a shadow 400 ft long . Find the angle of elevation
of the sun.
C

C
165 m

150 ft

Let :

400
ft. 52
Horizontal
A

tan A

tan A

be the angle of elevation

52

tan -1

20 33 22

EXAMPLE 17 : The angles of elevation of the peak of a mountain from two points A and
B on the ground, were found to be 0.95 & 0.92, respectively. If point B is
200 ft. farther from A, find the height of the mountain.
C

mountain
0.95

0.92
A

B
200 ft

Solution:
Let :

CD

be the height of the mountain

To solve for CD:

Consider the triangle ADC and triangle BDC
tan 0.95 =

CD = AD tan 0.95 ---- Eq. 1

tan 0.92

CD = (AD +200) tan 0.92 ---- Eq. 2

Since CD are equal in the triangle ADC and triangle BDC :

AD tan 0.95 = (AD + 200) tan 0.92
Simplifying,
AD tan 0.95 AD tan 0.92

200 tan 0.92

AD (tan 0.95 tan 0.92)

200 tan 0.92

18 - H

Substitute to Eq. (1)

CD = AD tan 0.95

CD = 4,315.012 ft
Example 18: From the top of the building, 18 meters high, the angle Hof depression of the
top of a flagpole is 21. From the bottom of the building, the angle of
elevation of the top of the flagpole is 48. Find the height of the flagpole
and its distance from the building.
Solution:
Let : H = be the height of the flagpole
D = be its distance from the building.

18 m

Consider the two right triangles:

48
21

21

With respect to 21: tan 21

With respect to 48: tan 48

------------

Eq. 1
D

-----------

Eq.

Since D in both triangles are equal:

D1

D2

Simplifying:
H tan 21
H(tan 21 + tan 48)

=
=

(18 H) tan 48
18tan48

Solving for H :
H =
13.3766 m or

13 m

Substitute H = 13.3766 to Eq. 2

D =

= 12.0444 or 12 m

Check:

0.38386

= 0.38386

3.3 BEARING OF A LINE (OR A POINT) - is the angle from North or South to the line
of sight of a point of reference. Figure 6 shows two mutually perpendicular lines, in
which NS represents the north-south medium and EW the east-west parallel.
N
B

25

48

O
80

40
C

S
Fig.6

Bearing of OA

: N 48 E

(Read as : 48 E of North)

Bearing of OB

: N 25 W (Read as : North 25 towards West)

Bearing of OC

: S 40 W (Read as : 40 West of South)

Bearing of OD

: S 80 E

(Read as : South 80 towards East)

3.4 COURSE OF A SHIP OR PLANE the angle drawn from the North line,
clockwise direction, to the line of travel of the carrier (Figure 7).

55
A
C

N
180
340

Course of A : 55
Course of B : 180
B

Course of C : 340

X
W

Fig. 7

EXAMPLE 19: A ship is sailing 150 miles in the direction N 3810 W. How far north
and how far west of the starting point is it?

N
Ship

sin 3810 =

x = 150 sin3810
50 miles

x = 92.69 miles West

0

38 10'
cos 3810 =

y = 150 cos 3810

y = 117.93 miles North

Example 20: An airplane flies 30 miles due North and then proceeds to westward at a
speed of 15 miles per hour. Find the bearing of the plane and its distance
from the starting point 6 hours after it turned west ward.
Solution :
E
The plane started at A, turned west at B and reached C 6 hours after it turned at B.
N
C

30 mi

Let: AC = be the distance of the plane from the starting point.

BC = RT =

(6 hr)

BC = 90 miles
To solve for the bearing:
tan =

tan =

then, = tan-1

= 7134

4330

Bearing is N 7134 W
C

Solve for AC : cos =

AC =

AC = 94.876 miles
Check : tan 7134

?=

3 = 3 OK
Example 21: A ship is sailing due West when a lighthouse is observed bearing N 6520
W. After travelling 2200, the light house is bearing N 4630 W. If the
course is continued, how close will the ship approach the light?

244

Solution:

46 30 '

2200 ft

Let :

= be the closest distance of the ship

L=

be the position of the light

A=

be the first position of the ship

B=

be the second position

C=

be the position nearest L.

In triangle ACL:

tan 2440 =

CL = (BC+2200) tan 2440

In triangle BCL:

---------------Eq.

tan 4330 =

CL = BC tan 4334

---------------Eq. 2

Since CL are equal in both triangles:

BC tan 4330 = (BC + 2200) tan 2440
BC tan 4330 BC tan 2440 = 2200 tan 2440
BC ( tan 4330 - BC tan 2440 ) = 2200 tan 2440
BC =
-------------------

Eq. 3

Substitute Eq.3 to Eq. 2:

CL =

CL = 1,957.7973 ft
Check :
?
tan 2440 =

tan 4330

0.4592

0.4592