In ΔABC, if A and B are complementary angles such that tan2A + 16 tan2 B + tan A + 4 tan B = 12 and c = 8, then Column I Column II (A) length of the altitude from the vertex C to the side AB is (P) 2 (B) area of the ΔABC is (Q) \frac{16}{5} (C) the ratio \frac{sinA}{sinB} equals (R) 4 (D) length of the median from vertex C to the side AB is (S) \frac{64}{5}

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

In ΔABC, if A and B are complementary angles such that tan²A + 16 tan² B + tan A + 4 tan B = 12 and c = 8, then

Column I	Column II
(A) length of the altitude from the vertex C to the side AB is	(P) 2
(B) area of the ΔABC is	(Q) $\frac{16}{5}$
(C) the ratio $\frac{sinA}{sinB}$ equals	(R) 4
(D) length of the median from vertex C to the side AB is	(S) $\frac{64}{5}$

A

A → Q; B → R; C → P; D → S

B

A → Q; B → S; C → P; D → R

C

A → S; B → Q; C → P; D → R

D

A → R; B → S; C → P; D → Q

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tan²A + 16 tan² B + tan A + 4 tan B = 12

tan²A + 16 cot² A + tan A + 4 cot A = 12

(tan A – 4 cot A)² + ${(\sqrt{tanA} - 2 \sqrt{cotA})}^{2}$ = 0

tan A = 4 cot A and $\sqrt{\tan A} = 2 \sqrt{\cot A}$ = 2

⇒ tan A = 2 and tan B = $\frac{1}{2}$

(A) tan A = $\frac{p}{x}$ = 2 ⇒ p = 2x

$\tan B = \frac{p}{8 - x} = \frac{1}{2}$ ⇒ 2p = 8 – x

⇒ $p = \frac{16}{5}, x = \frac{32}{5}$

(B) $Δ = \frac{1}{2} \times 8 \times \frac{16}{5} = \frac{64}{5}$

(C) $\frac{\sin A}{\sin B} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2$

(D) Median CE = l = 4

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