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UPSC CDS II 2024 Math Paper

Question Numbers: 51-53

Consider the following for the items that follow:

Let ABC be a triangle right-angled at B. Let P be the point on BC such that BP = PC. If A

B = 10

cm, ∠BA

P = 45

° and ∠CAP = θ

(usetan(\alpha +\beta )=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta })

Question : 51 of 100

Marks: +1, -0

What is tanθ equal to?

Solution:

Concept:

Use the tangent addition formula and the given geometric relationships to find

\tan \theta

Explanation:

\triangle \mathrm{ABP}

\angle \mathrm{BAP} = 45^\circ

and

\mathrm{AB} = 10\ \mathrm{cm}

\mathrm{BP} = \mathrm{CP}

\tan 45^\circ = \frac{\text{BP}}{\text{AB}} \implies 1 = \frac{\text{BP}}{\text{AB}} \implies \text{BP} = \text{AB}

\triangle \mathrm{BAC}

\angle \mathrm{BAC} = 45^\circ + \theta

\tan(45^\circ + \theta) = \frac{\mathrm{BC}}{\mathrm{AB}} = \frac{\mathrm{BP} + \mathrm{PC}}{\mathrm{AB}} = \frac{2 \cdot \mathrm{BP}}{\mathrm{AB}} = \frac{2 \cdot \mathrm{AB}}{\mathrm{AB}} = 2

Using the tangent addition formula:

\tan(45^\circ + \theta) = \frac{\tan 45^\circ + \tan \theta}{1 - \tan 45^\circ \cdot \tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta}

Set equal to

2

2 = \frac{1 + \tan \theta}{1 - \tan \theta}

2(1 - \tan \theta) = 1 + \tan \theta \implies 2 - 2\tan\theta = 1 + \tan\theta \implies 1 = 3\tan\theta \implies \tan\theta = \frac{1}{3}

Answer:

\frac{1}{3}

(Option B).

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