Mathematics

In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:
$\Big(\dfrac{4}{3}\Big)$
$\Big(\dfrac{5}{3}\Big)$
$\Big(\dfrac{13}{3}\Big)$
$\Big(\dfrac{14}{3}\Big)$

Trigonometric Identities

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Answer

ST = PS − RQ = 14 − 5 = 9 cm

In ΔSTR,

$TR = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.$

We know that,

$\tan \alpha = \dfrac{\text{opposite}}{\text{adjacent}} \\[1em] \tan \alpha = \dfrac{TR}{ST} \\[1em] \tan \alpha = \dfrac{12}{9} \\[1em] \tan \alpha = \dfrac{4}{3}.$

Hence, option 1 is the correct option.

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Mathematics

In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:
$\Big(\dfrac{4}{3}\Big)$
$\Big(\dfrac{5}{3}\Big)$
$\Big(\dfrac{13}{3}\Big)$
$\Big(\dfrac{14}{3}\Big)$

Trigonometric Identities

Answer

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Mathematics

In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:(43)\Big(\dfrac{4}{3}\Big)(34​)(53)\Big(\dfrac{5}{3}\Big)(35​)(133)\Big(\dfrac{13}{3}\Big)(313​)(143)\Big(\dfrac{14}{3}\Big)(314​)

Trigonometric Identities

Answer

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In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:
$\Big(\dfrac{4}{3}\Big)$
$\Big(\dfrac{5}{3}\Big)$
$\Big(\dfrac{13}{3}\Big)$
$\Big(\dfrac{14}{3}\Big)$

If 3 cos θ = 1, then the value of cosec θ is:
$2\sqrt{2}$
$\Big(\dfrac{3}{2\sqrt{2}}\Big)$
$\Big(\dfrac{2\sqrt{3}}{3}\Big)$
$\Big(\dfrac{4}{3\sqrt{2}}\Big)$

If x cos A = 1 and tan A = y, then x² − y² is equal to:
0
1
−tan A
tan A

The value of (1 + tan θ + sec θ)(1 + cot θ − cosec θ) is:
−4
−1
1
2

If sin θ = $\Big(\dfrac{1}{2}\Big)$ , then the value of $\Big(\dfrac{1}{5} \cot^2\theta + \dfrac{1}{5}\Big)$ is:
$\Big(\dfrac{1}{5}\Big)$
$\Big(\dfrac{4}{5}\Big)$
$\Big(\dfrac{1}{125}\Big)$
25