Question

Sin 60* . Sec 30* plz help to solve this question
i will mark as brainlist
trigonometry chapter​

Answer 1

Step-by-step explanation:

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

Answer 2

HOPE IT HELPS YOU!!✅✔️