Question

Find sin60 and sec30 geometrically

Answer 1

find sin 60.

Let Δ ABC be an equilateral triangle.
∠A=∠B=∠C= 60°
lets draw AD⊥BC
Now in ΔABD and ΔACD
AD=AD;
∠ADB =∠ADC  (=90)
AB=AC  (sides of equilateral triangle)
∴ΔABD  is congruent to ΔACD.
since sides of equilateral triangle are equal, we can write :  c=2a.  ----->(1)
 Using Pythagorean theorem in ΔABD
c²=a²+b²;
(2a)²=a²+b²;
b=√((2a)²-a²)=√3a
b=√3a    ----------->(2)

now in Δ ABD
sin 60= perpendicular / hypoteneous =AD/AB= b/c =(√3a) / 2a= √3/2
So sin 60= √3 /2
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Find sec 30°

In the same triangle ABC
cos 30=AD/AB
         =  b/c
from eq(1) and (2)
   cos 30= (√3a)/2a
so cos 30=√3/2;

SEC 30°= 1/ cos 30°= 2/√3.

So sec 30° = 2/√3

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Answer 2

Let Δ ABC be an equilateral triangle.

∠A=∠B=∠C= 60° (ANGLE SUM PROPERTY)

lets draw AD⊥BC

Now in ΔABD and ΔACD

AD=AD; (COMMON)

∠ADB =∠ADC  (=90)

AB=AC  (sides of equilateral triangle)

∴ΔABD  is congruent to ΔACD.

since sides of equilateral triangle are equal, we can write :  c=2a.  ----->(1)

Using Pythagorean theorem in ΔABD

c²=a²+b²;

(2a)²=a²+b²;

b=√((2a)²-a²)=√3a

b=√3a    ----------->(2)

now in Δ ABD

sin 60= perpendicular / hypotenuse =AD/AB= b/c =(√3a) / 2a= √3/2

So sin 60= $\sqrt{3} /2$

HOPE YOU UNDERSTOOD THANKS