Find the value of sec 45 geometrically
find the value of sin 30 geometrically

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Solution

First of all take a triangle ABC for which angle A=angle B=45
then angle C=90
We know it is isosceles rightangled triangle.
Let give value for equal sides 1 cm.Then using Pythagoras theorm,
hypotenyse=√(1^2)+(1^2)
=√2
Cos 45=adjacent side÷hypotenuse=1÷√2
So sec 45 =1÷(1÷√2)
=√2

similarly consider another triangle with angles 30,60,90...we know sides opposite to these angles are in ratio 1:√3:2 respectively.
we consider length of side opposite to 30 is 1 cm.then side opposite to 60 is √3cm and side opposite to 90 is 2cm which is the hypotenuse of this triangle.

sin 30 =opposite side of angle 30÷hypotenuse
=1÷2

so answers all get correctly. write once to paper for understanding.

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Similar questions

Q. Find the value of sin 30° geometrically.

Q. Find the value of

\frac{sin 30^{\circ} + tan 45^{\circ} - csc 60^{\circ}}{cot 45^{\circ} + cos 60^{\circ} - sec 30^{\circ}}

Q. The value of expression

\frac{sin 30^{\circ} + tan 45^{\circ} - sec 60^{\circ}}{cosec 30^{\circ} - cot 45^{\circ} - cos 60^{\circ}}

Find the value of

$\frac{1 + s i n 30^{\circ}}{c o s 30^{\circ}} + \frac{c o s 30^{\circ}}{1 + s i n 30^{\circ}}$

Show that the value of $\frac{s i n 45^{\circ} - s i n 30^{\circ}}{c o s 45^{\circ} + c o s 60^{\circ}}$ and $\frac{s e c 45^{\circ} - t a n 45^{\circ}}{c o s e c 45^{\circ} + c o t 45^{\circ}}$ are equal

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Find the value of sec 45 geometrically find the value of sin 30 geometrically

Find the value of sec 45 geometrically
find the value of sin 30 geometrically