Question

The length of side AC of the triangle △ABC when ∠A = 65°, AB = 7cm, and BC = 12cm. *

Answer 1

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Answer 2

Given :- In ∆ABC, ∠A = 65°, AB = 7cm, and BC = 12cm.

To Find :- AC = ?

Formula used :- Cosine rule :-

• a² = b² + c² - 2bc•cos A

Solution :-

from given values we get,

→ a = 12 cm

→ b = Let x cm

→ c = 7 cm

→ ∠A = 65°

So, putting all values in cosine rule formula we get,

→ 12² = x² + 7² - 2 * x * 7 * cos 65°

→ 144 - 49 = x² - 14x * cos 65°

→ 95 = x² - 14x * 0.423

→ x² - 5.92x - 95 = 0

→ x² - (592x/100) - 95 = 0

→ x² - (148x/25) - 95 = 0

→ 25x² - 148x - 2375 = 0

Solving this quadratic equation now by using sridharacharya formula for Solving quadratic equation ax² +bx + c = 0 ;

• x = [ -b±√(b²-4ac) / 2a ]

putting values we get,

→ x = [-(-148) ± √{(-148)² - 4 * 25 * (-2375)}]/2*25

→ x = [148 ± √(21904 + 237500)] / 50

→ x = [148 ± √259404]/50

→ x = [148 ± √(4 * 64851)]/50

→ x = 2[74 ± √(64851)]/50

→ x = (74 ± √64851)/25

→ x = (74 + √64851)/25 or (74 - √64851)/25

since length of sides of ∆ cant be negative . Taking positive value we get,

→ x = (74 + √64851)/25

→ x = (74 + 254.6)/25

→ x = (328.6/25)

→ x = 13.14 cm (Ans.)

Hence, The length of side AC of the triangle is equal to 13.14 cm (approximately).

Learn more :-

In the figure ∠ MNP = 90°, ∠ MQN = 90°, , MQ = 12 , QP = 3 then find NQ .

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