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8. Introduction to Trigonometry
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यदि $\sin 3 A =\cos \left( A -26^{\circ}\right)$ हो, जहाँ, $3 A$ एक न्यून कोण है तो $A$ का मान जात कीजिए।
A
$29$
B
$32$
C
$20$
D
$25$
Solution
We are given that $\sin 3 A =\cos \left( A -26^{\circ}\right)$ ……..$(1)$
Since,$\sin 3 A =\cos \left(90^{\circ}-3 A \right),$ we can write $(1)$ as
$\cos \left(90^{\circ}-3 A \right)=\cos \left( A -26^{\circ}\right)$
Since,$90^{\circ}-3 A$ and $A -26^{\circ}$ are both acute angles, therefore,
$90^{\circ}-3 A = A -26^{\circ}$
Which Gives,$A=29^{\circ}$
Standard 10
Mathematics
