C-09-Q188mediumsingle_mcq

Solve: $\sin\theta + \cos\theta = 1$ , $0^\circ \le \theta \le 90^\circ$ . $\theta = ?$

aonly $0^\circ$
bonly $90^\circ$
c $0^\circ$ or $90^\circ$
d $30^\circ$

ব্যাখ্যা

Squaring gives $1+2\sin\theta\cos\theta=1$ , so $\sin\theta\cos\theta=0$ . Within $[0^\circ,90^\circ]$ this holds at both endpoints, giving $\theta=0^\circ$ or $90^\circ$ .

এই অধ্যায়ের আরও প্রশ্ন

C-09-Q189Solve: , acute.
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C-09-Q192With the same data,
C-09-Q193If and with acute, then
C-09-Q194With the same data,
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Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

Solve: $\sin\theta + \cos\theta = 1$ , $0^\circ \le \theta \le 90^\circ$ . $\theta = ?$

aonly

0^\circ

bonly

90^\circ

0^\circ

90^\circ

30^\circ

ব্যাখ্যা

Squaring gives $1+2\sin\theta\cos\theta=1$ , so $\sin\theta\cos\theta=0$ . Within $[0^\circ,90^\circ]$ this holds at both endpoints, giving $\theta=0^\circ$ or $90^\circ$ .

Solve: sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1sinθ+cosθ=1, 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ0∘≤θ≤90∘. θ=?\theta = ?θ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve: sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1sinθ+cosθ=1, 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ0∘≤θ≤90∘. θ=?\theta = ?θ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve: $\sin\theta + \cos\theta = 1$ , $0^\circ \le \theta \le 90^\circ$ . $\theta = ?$

Solve: $\sin\theta + \cos\theta = 1$ , $0^\circ \le \theta \le 90^\circ$ . $\theta = ?$