**Problems on Triangle in Trigonometry for Class 11 :**

Here we are going to see some example problems on triangle in trigonometry for class 11.

**Question 1 :**

In any triangle ABC, prove that the area triangle = b^{2} + c^{2} − a^{2}/4 cotA.

**Solution :**

Cosine formula :

cos A = (b^{2} + c^{2} - a^{2}) / 2bc

Given that :

= b^{2} + c^{2} − a^{2}/4 (cos A/sin A)

= (b^{2} + c^{2} − a^{2}) sin A/4 cos A ---(1)

From cosine formula, we have

2bc cos A = b^{2} + c^{2} - a^{2}

By applying the value of b^{2} + c^{2} - a^{2 }in (1)

= (2bc cos A) sin A/4 cos A

= (bc sin A)/2

= (1/2) bc sin A

= Area of triangle

Hence proved.

**Question 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.

**Question 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.

**Question 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)

After having gone through the stuff given above, we hope that the students would have understood, "Problems on Triangle in Trigonometry for Class 11"

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