In a triangle ABC- a - 6- b - 3 and tanA-B2-13- Area of triangle ABC - sq- units-

Using Napier's rule, we can say that tan nbsp;(A B2)=a ba+bcot nbsp;C2⇒13=6 36+3cot nbsp;C2⇒cot nbsp;C2=1⇒C2=π4⇒c=π2∴Area nbsp;of nbsp;∆ nbsp;ABC ...

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Byju's Answer

Standard XI

Mathematics

Napier’s Analogy

In a triangle...

Question

In a triangle ABC, a = 6, b = 3 and $tan (\frac{A - B}{2}) = \frac{1}{3}$ . Area of triangle ABC = sq. units.

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Solution

The correct option is D 9
Using Napier's rule, we can say that-
$tan (\frac{A - B}{2}) = \frac{a - b}{a + b} cot \frac{C}{2} \Rightarrow \frac{1}{3} = \frac{6 - 3}{6 + 3} cot \frac{C}{2} \Rightarrow cot \frac{C}{2} = 1 \Rightarrow \frac{C}{2} = \frac{π}{4} \Rightarrow c = \frac{π}{2} ∴ Area of Δ ABC = \frac{1}{2} \times 6 \times 3 \times sin \frac{π}{2} = 9 sq . units$

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In a triangle ABC, a = 6, b = 3 and tan(A−B2)=13. Area of triangle ABC = sq. units.

The correct option is D 9Using Napier's rule, we can say that- tan (A−B2)=a−ba+bcot C2⇒13=6−36+3cot C2⇒cot C2=1⇒C2=π4⇒c=π2∴Area of Δ ABC=12×6×3×sin π2=9 sq. units

In a triangle ABC, a = 6, b = 3 and $tan (\frac{A - B}{2}) = \frac{1}{3}$ . Area of triangle ABC = sq. units.