Question

ABC is a triangle right angle at A if AB = 12 cm and Ac = 5 cm, find BC

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Answer 1

Answer:

use Pythagoras theorem

Answer 2

Answer:

$The value of \bold{\sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}}sinA=135, TanA=125,sinC=1312,cotC=135. \\ </p><p></p><p>Solution: \\ </p><p></p><p>The \: triangle \: ABC \: is \: drawn \: below \: which \: is   \\ </p><p></p><p>In \: the \: triangle \: ABC \: angle \: B \: is \: 90 \: degree. The \: “length \: of \: the \: side \: AB” = 12cm \: and \: the \: “length \: of \: the \: side \: BC” = 5cm. \\ </p><p></p><p>Now \: to \: find \: the \: “length \: of \: side \: AC” \: we \: use \: Pythagoras \: theorem \: we \: get  \: A B^{2}+B C^{2}=A C^{2}AB2+BC2=AC2 \\ </p><p></p><p>\sqrt{A B^{2}+B C^{2}}=A CAB2+BC2=AC \\ </p><p></p><p>=\sqrt{12^{2}+5^{2}}=122+52 \\ </p><p></p><p>=\sqrt{144+25}=13=144+25=13 \\ </p><p></p><p>Now \: the \: question \: says \: to \: find \: 1) Sin \: A \: and \: Tan \: A, \: 2) Sin \: C \: and \: Cot \: C. \\ </p><p></p><p>So \: using \: the \: formula, \: we \: get \\ </p><p></p><p>\sin A=\frac{\text {height}}{\text {hypotenuse}}=\frac{B C}{A C}=\frac{5}{13}sinA=hypotenuseheight=ACBC=135 \\ </p><p></p><p>\tan A=\frac{\text {height}}{\text {base}}=\frac{B C}{A B}=\frac{5}{12}tanA=baseheight=ABBC=125 \\ </p><p></p><p>\sin C=\frac{\text {height}}{\text {hypotenuse}}=\frac{A B}{A C}=\frac{12}{13}sinC=hypotenuseheight=ACAB=1312 \\ </p><p></p><p>\cot C=\frac{\text {base}}{\text {height}}=\frac{B C}{A B}=\frac{5}{13}cotC=heightbase=ABBC=135 \\ </p><p></p><p>Therefore, the value of \sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}sinA=135, TanA=125,sinC=1312,cotC=135 \\ </p><p></p><p>$