Question

Guinen that tan a -5/12 and angle
q is a acute angle find sin q and col q​

Answer 1

$Given:- \\ \tan(A) \: = \: \frac{ - 5}{12} \\ To \: Find:- \\ \sin(Q) \\ \cos(Q) \\ \tan(Q) \\ \sec(Q) \\ \csc(Q) \\ \cot(Q) \\ Let \: a \: triangle \: ∆ABQ \: right \: angled \: at \: B \\ therefore, \: QB \: = \: 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AB \: = \: - 5 \\ using \: pythagoras \: theo. \: we \: get, \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AQ \: = \: 13 \\ now \\ \sin(Q) \: = \: \frac{ - 5}{13} \\ \cos(Q) \: = \: \frac{12}{13} \\ \tan(Q) \: = \: - \frac{12}{5} \\ \sec(Q) \: = \: \frac{13}{12} \\ \csc(Q) \: = \: \frac{13}{ - 5} \\ \cot(Q) \: = \: \frac{ - 5}{12} \\ \\ \\ \\ hope \: it \: helps.................. \\ plz \: mark \: as \: brainliest.....................$

Answer 2

Answer:

Sin Q = 5/13, Cos Q = 12/13

Step-by-step explanation:

Given : Tan Q = 5/12

Tan Q = PR/QR = 5x/12x

If ΔPQR is taken with the right angle being R, (as shown)

then using Pythagoras theorem, we get PQ² = PR² + QR²

PQ² = (5x)² + (12x)² = 25x² + 144x² = 169x²

∴ PQ = 13x

Now, Sin Q = PR/PQ = 5x/13x

Sin Q = 5/13

Cos Q = QR/PQ = 12x/13x

Cos Q = 12/13