Question

ம‌தி‌ப்‌‌பிடுக
(i) (sin49°)/(cos41°) (ii) (sec63°)/(csc27°)

Answer 1

Step-by-step explanation:

Hi friend

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Your answer

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In ∆ ABC , We have,

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AB = AC

so, ∆ABC is an isosceles triangle.

Then,

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Angle B = Angle C

Angle A = 50° (given)

Angle B = Angle C = (180° - 50°)/2

= 130°/2

= 65°

HOPE IT HELPS

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Answer 2

விளக்கம்:

$\begin{aligned}&\text { (i) } \frac{\sin 49^{\circ}}{\cos 41^{\circ}}\\&\sin 49^{\circ}=\sin \left(90^{\circ}-41^{\circ}\right)\end{aligned}$

           $=\cos 41^{\circ}$

ஏனெனில் $49^{\circ}+41^{\circ}=90^{\circ}$

$\sin 49^{\circ}=\cos 41^{\circ}$ எனப் பிரதியிட

$\frac{\cos 41^{\circ}}{\cos 41^{\circ}}=1$

$\text { (ii) } \frac{\sec 63^{\circ}}{\csc 27^{\circ}}$ ன் மதிப்பு

$\sec 63^{\circ}=\sec \left(90^{\circ}-27^{\circ}\right)$

$=\csc 27^{\circ}$ எனப் பிரதியிட

$\frac{\sec 63^{\circ}}{\csc 27^{\circ}}=\frac{\csc 27^{\circ}}{\csc 27^{\circ}}$

            = 1