SKETCHING A TRIANGLE AND DETERMINE THE MEASURE OF INDICATED SIDE

Sketch each triangle. Determine the measure of the indicated side.

Example 1 :

In triangle ABC, ∠A = 57°, ∠B = 73°, and AB = 24 cm. Find the length of AC.

Solution :

AB  =  c  =  24 cm, BC  =  a, AC  =  b, <A  =  57 and <B  =  73

In triangle ABC,

<A + <B + <C  =  180

57 + 73 + <C  =  180

<C  =  180 - 130  =  50

Using sin formula,

a/sin A  =  b/sin B  =  c/sin C

a/sin 57  =  b/sin 73  =  24/sin 50

We have to solve for b.

Equating 1 and 3, we get

b/sin 73  =  24/sin 50

b/0.9563   =  24/0.7660

b  =  31.33(0.9563)

b  =  29.96

b  =  30 cm (approximately)

Hence the indicated side is 30 cm.

Example 2 :

In triangle ABC, ∠B = 38°, ∠C = 56°, and BC = 63 cm. Find the length of AB.

Solution :

AB  =  c, BC  =  a = 63, AC  =  b, <C  =  56 and <B  =  38

In triangle ABC,

<A + <B + <C  =  180

<A + 38 + 56  =  180

<A  =  180 - 94  =  86

Using sin formula,

a/sin A  =  b/sin B  =  c/sin C

63/sin 86  =  b/sin 38  =  c/sin 56

We have to solve for c.

Equating 1 and 3, we get

63/sin 86  =  c/sin 56

63/0.9975   =  c/0.8290

63.15  =  c/0.8290

c  =  63.15(0.8290)

c  =  52.35

c  =  52.4 cm (approximately)

Hence the indicated side is 52.4 cm.

Example 3 :

In triangle ABC, ∠A = 50°, ∠B = 50°, and AC = 27 m. Find the length of AB.

Solution :

AB  =  c, BC  =  a = 27 m,  AC  =  b = 27 m, <A  =  50 and <B  =  50

In triangle ABC,

<A + <B + <C  =  180

50 + 50 + <C  =  180

<C  =  180 - 100  =  80

Using sin formula,

a/sin A  =  b/sin B  =  c/sin C

27/sin 50  =  27/sin 50  =  c/sin 80

We have to solve for c.

Equating 1 and 3, we get

27/sin 50  =  c/sin 80

27/0.7660   =  c/0.9848

35.24  =  c/0.9848

c  = 35.24(0.9848)

c  =  34.70 m

Hence the indicated side is 34.7 m.

Find the length of the remaining side in the given triangle:

Example 4 :

Solution :

The side which is opposite to <A = a, <B = b and <C = c

AB = c = 15 cm, BC = a = ?,  AC = b = 21 cm, <CAB = 105

Since we know the two sides and one angle measure, we have to use the cosine rule

a² = b² + c² - 2bc cos A

a² = 21² + 15² - 2(21)(15) cos 105

= 441 + 225 - 630(-0.25)

= 666 + 157.5

a² = 823.5

a = √823.5

a = 28.69 cm

Example 5 :

Solution :

The side which is opposite to <P = p, <Q = q and <R = r

QR = p = 4.8 km, PR = q = 6.3 km,  PQ = r = ?, <QRP = 32

Since we know the two sides and one angle measure, we have to use the cosine rule

r² = p² + q² - 2pq cos R

r² = (4.8)² + (6.3)² - 2(4.8)(6.3) cos 32

= 23.04 + 39.69 - 60.48 cos 32

= 62.73 - 60.48(0.84)

= 62.73 - 50.80

= 11.93

r = √11.93

r = 3.45 km

Example 6 :

Solution :

The side which is opposite to <K = k, <L = l and <M = m

LM = k = 14.8 m, KM = l = ?, LK = m = 6.2 m, <KLM = ?

Since we know the two sides and one angle measure, we have to use the cosine rule

l² = k² + m² - 2km cos L

l² = (14.8)² + (6.2)² - 2(14.8)(6.2) cos 72

= 219.04 + 38.44 - 183.52 cos 72

= 257.48 - 183.52(0.309)

= 257.48 - 56.70

= 200.78

l = √200.78

l = 14.1 m

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