In a right-angled triangle PQR, where \angle Q = 90^\circ and \tan P = \frac{5}{12}, what … — JAMB UTME

In a right-angled triangle $PQR$ , where $\angle Q = 90^\circ$ and $\tan P = \frac{5}{12}$ , what is the exact value of $\sin P + \cos P$ ?

\frac{17}{13}

Using

\tan P = \frac{5}{12}

, the triangle has opposite side

5

, adjacent side

12

, and hypotenuse

\sqrt{5^2+12^2} = 13

. Therefore,

\sin P = \frac{5}{13}

and

\cos P = \frac{12}{13}

, giving a sum of

\frac{17}{13}

Identify the side lengths of the right triangle

PQR

using the given tangent ratio.

Opposite side to

P = 5

, adjacent side to

P = 12

By definition,

\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12}

Calculate the length of the hypotenuse

PR

using the Pythagorean theorem.

Hypotenuse

PR = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

In a right triangle,

\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2

Determine

\sin P

and

\cos P

and compute their sum.

\sin P = \frac{5}{13}

\cos P = \frac{12}{13}

, so

\sin P + \cos P = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}

\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}}

and

\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}}

Basic Trigonometric Ratios and Pythagorean Triples

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