PROBLEMS ON TRIANGLE IN TRIGONOMETRY FOR CLASS 11

Problem 1 :

In any triangle ABC, prove that the area triangle is

b² + c² − a²/4 cotA

Solution :

Cosine formula :

cos A  =  (b² + c² - a²) / 2bc

Given that :

=  b² + c² − a²/4 (cos A/sin A)

=  (b² + c² − a²) sin A/4 cos A  ---(1)

From cosine formula, we have

2bc cos A  =  b² + c² - a²

By applying the value of b² + c² - a² in (1)

  =  (2bc cos A) sin A/4 cos A

  =  (bc sin A)/2

  =  (1/2) bc sin A

  =  Area of triangle

Hence proved.

Problem 2 :

In a triangle ABC, if a = 12 cm, b = 8 cm and C = 30°, then show that its area is 24 sq.cm.

Solution :

Area of triangle  =  (1/2) ab sin C

  =  (1/2) 12(8) sin 30°

  =  48 (1/2)

  =  24 square units.

Hence it is proved.

Problem 3 : 

In a triangle ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.

Solution :

Since the length of all sides are different, we have to use heron's formula to find the area.

a = 18 cm, b = 24 cm and c = 30 cm

s  =  (a + b + c)/2

  =  (18 + 24 + 30)/2

  =  72/2

 s = 36

s - a = 36-18  =  18

s - b = 36-24  =  12

s - c = 36-30  =  6

Area of triangle  = √s(s-a)(s-b)(s-c)

  = √36(18)(12)(6)

  =  216 square units.

Hence it is proved.

Problem 4 :

Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5km apart, find the distance of the intruder from B.

Solution :

<A  =  30

<B + 45  =  180

<B  =  180 - 45  =  135

<A + <B  + <C  =  180

30 + 135 + <C  =  180

<C  =  180 - 165

<C  =  15

BC  =  a, AC  =  b and AB  =  c

Sine formula :

a/sin A  =  b/sin B  =  c/sin C

a/sin 30  =  b/sin 135  =  5/sin 15

 a/(1/2)  =  5/sin (45-30)

2a  =  5/sin (45-30)

sin (45-30) = sin 45 cos 30 - cos 45 sin 30

    =  (1/√2)(√3/2) -  (1/√2)(1/2)

   =  √3/2√2 - 1/2√2

   =  (√3 - 1)/2√2

2a  =  5/[(√3 - 1)/2√2]

2a  =  10√2/(√3 - 1)

a  =  5√2/(√3 - 1)

Problem 5 :

A stationary surveillance camera is set up to monitor activity in the parking lot of a shopping mall. If the  camera has a 38° field of vision, how many square feet of the parking lot can it tape using the given dimensions?

Solution :

We start with sketching an appropriate diagram. Assume the notation as in figure

From the sine formula for area of a triangle, we have

PRS = (1/2) ∙ 110 ∙ 225 sin 38°

= 7619 ft²

The surveillance camera monitors approximately 7619  square feet of the parking lot.

Problem 6 :

A New York City developer wants to build condominiums on the triangular lot formed by Greenwich, Watts, and Canal Streets. How many square meters does the  developer have to work with if the frontage along each street is approximately 34.1 m, 43.5 m, and 62.4 m, respectively?

Solution :

To find the area of the triangular lot with given sides, we would like to use Heron’s Formula. For this reason, we first calculate the semi perimeter

s = (34.1 + 43.5 + 62.4)/2

= 70

Area of the triangle

= √70(70 - 34.1) (70 - 43.5)(70 - 62.4)

= √(70 x 35.9 x 26.5 x 7.6)

= √506118.2

= 711 m²

Problem 7 :

To find the distance AB across a river, a surveyor laid off a distance BC = 354 meters on one side of the river, as shown in the accompanying figure. It is found that ∠B = 112°10′ and ∠C = 15°20′. Find the distance AB.

Solution :

a/sin A = b/sin B = c/sin C

A = 180 - (15°20′ + 112°10′)

= 180 - 127°30′

= 52°30′

354/sin 52°30′ =  AC/sin 112°10′ = AB/sin 15°20′

354/sin 52°30′ = AB/sin 15°20′

AB = 354 (sin 15°20′)/sin 52°30′

= 354(0.2644)/0.7934

= 93.59/0.7934

= 117.9

So, the required distance of AB is 118 m approximately.

Problem 8 :

To determine the distance RS across a deep canyon (see
the accompanying figure), Peter lays off a distance 𝑇𝑅 = 480 meters. Then he finds that ∠𝑇 = 32° and ∠𝑅 = 102°.
Find the distance RS

Solution :

a/sin A = b/sin B = c/sin C

S = 180 - (102 + 32′)

= 180 - 134

= 46°

ST/sin 102° = SR/sin 32° = RT/sin 46°

SR/sin 32° = RT/sin 46°

SR/sin 32° = 480/sin 46°

SR = 480 (sin 32)/sin 46

= 480 (0.5299)/0.719

= 353.7

Approximately 354 m.

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