Given thatsin A - 1-2 and cos B - 1-2- then find the value of A - B-Here- 0 - A - B - 90- -2 Marks-

Here, 0 lt; A + B ≤ 90^∘ nbsp; sin A = 1/2 (given) ⇒ nbsp;sin A = sin 30^∘ =1/2 ⇒ A = 30^∘ nbsp; nbsp; nbsp;[1 Mark] cos B = nbsp;1/√(2 ...

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Standard IX

Mathematics

Complementary Trigonometric Ratios

Given thatsin...

Question

Given that $sin A = \frac{1}{2}$ and $cos B = \frac{1}{\sqrt{2}}$ , then find the value of $(A + B)$ .

$(H e r e, 0 < A + B \leq 90^{\circ})$ [2 Marks]

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Solution

$H e r e, 0 < A + B \leq 90^{\circ}$

$s i n A = \frac{1}{2}$ (given)

$\Rightarrow s i n A = s i n 30^{\circ} = \frac{1}{2}$

$\Rightarrow A = 30^{\circ}$ [1 Mark]

$c o s B = \frac{1}{\sqrt{2}}$

$\Rightarrow c o s B = c o s 45^{\circ} = \frac{1}{\sqrt{2}}$

$\Rightarrow B = 45^{\circ}$

Then, $A + B = 75^{\circ}$ . [1 Mark]

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Similar questions

Q. Given that

s i n A = \frac{1}{2}

and

c o s B = \frac{1}{\sqrt{2}}

, then the value of

(A + B)

is ___________.
(Here, 0 < A + B ≤ 90°)

Given that sin A = $\frac{1}{2}$ and cos B = $\frac{1}{\sqrt{2}}$ , then the value of (A + B) is ____.

$(H e r e, 0 < A + B \leq 90^{\circ})$

If sin A = \frac{1}{2} and cos B = \frac{1}{\sqrt{2}}

,
then find the value of (A+B).

Given that sin $A = \frac{1}{2}$ and $c o s B = \frac{1}{\sqrt{2}}$ , then the value of (A + B) is:

(a) Prove that:
$(1 + t a n A)^{2} + (1 - t a n A)^{2} = 2 s e c^{2} A$

(b) If sin (A + B) = 1 and cos (A - B) = 1, find the values of A and B. Given $0 \leq 90^{\circ}$ and $A \geq B$ [6 MARKS]

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Given that sin A =12 and cos B =1√2, then find the value of (A+B). (Here, 0<A+B≤90∘) [2 Marks]

Here, 0<A+B≤90∘ sin A=12 (given) ⇒sin A=sin 30∘=12 ⇒A=30∘ [1 Mark] cos B=1√2 ⇒cos B=cos 45∘=1√2 ⇒B=45∘ Then, A+B=75∘. [1 Mark]

Given that $sin A = \frac{1}{2}$ and $cos B = \frac{1}{\sqrt{2}}$ , then find the value of $(A + B)$ .

$(H e r e, 0 < A + B \leq 90^{\circ})$ [2 Marks]