If in a triangle ABC- B-C- -u- u- -v- v- and A-C- 2u- u- where - u- - v- - then

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios of Complementary Angles

If in a trian...

Question

If in a triangle $A B C$ , $\to B C = \frac{\to u}{| \to u |} - \frac{\to v}{| \to v |}$ and $\to A C = \frac{2 \to u}{| \to u |}$ , where $| \to u | \neq | \to v |, t h e n$

1 + cos 2 A + c o s 2 B + c o s 2 C = 2

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sin A = cos B

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1 + cos 2 A + c o s 2 B + c o s 2 C = 0

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sin B = cos C

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Solution

The correct option is B $sin A = cos B$

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Similar questions

\to u, \to v, \to w

be such that

| \to u | = 1, | \to v | = 2, | \to w | = 3.

If the projection of

\to v

along

\to u

is equal to that of

\to w

along

\to u

and vectors

\to v, \to w

are perpendicular to each other then

| \to u - \to v + \to w |

equals

Q. Let

\to u, \to v, \to w

be such that

| \to u | = 1, | \to v | = 2, | \to w | = 3

. If the projection

\to v

along

\to u

is equal to that of

\to w

along

\to u

and

\to v, \to w

are perpendicular to each other then

| \to u - \to v + \to w |

equals:

Q. Let

\to u, \to v, \to w

be such that

| \to u | = 1, | \to v | = 2, | \to w | = 3

. If the projection of

\to v

along

\to u

is equal to that of

\to w

along

\to u

and

\to v, \to w

are perpendicular to each other, then

| \to u - \to v + \to w |

equals

Q. Let

\to u

\to v

\to w

be such that

| \to u |

= 1,

| \to v |

= 2,

| \to w |

= 3 . If the projection

\to v

along

\to u

is equal to that of

\to w

along

\to u

and

\to v

\to w

are perpendicular to each other , then

| \to u \to v + \to w |

equals

$I f \to u = \to a - \to b a n d \to v = \to a + \to b a n d | \to a | = | \to b | = 2, t h e n | \to u \times \to v | =$

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If in a triangle ABC, →BC=→u|→u|−→v|→v| and →AC=2→u|→u|, where |→u|≠|→v|,then

The correct option is B sin A=cos B

If in a triangle $A B C$ , $\to B C = \frac{\to u}{| \to u |} - \frac{\to v}{| \to v |}$ and $\to A C = \frac{2 \to u}{| \to u |}$ , where $| \to u | \neq | \to v |, t h e n$

The correct option is B $sin A = cos B$