Question

14. In triangle ABC, right
angled at B if sin A =>3/2.
find the value of sin C cos A-
cos C sin A​

Answer 1

Step-by-step explanation:

Think, It will be correct

Answer 2

$\large\mathrm\blue{✎~-2/4~✍}$

➙Question:

In triangle ABC, right

In triangle ABC, rightangled at B if sin A =>√3/2.

(Hypotenuse is the longest side.. it can be smaller!!! so your questions is wrong!! lets take sin = √3/2)

find the value of sin C cos A - cos C sin A

➙Solve:

sin A = $\large{\frac{side~opposite ~to~angle~A}{Hypotenuse}}$

sinA = $\frac{CB}{AC}~=~\frac{√3}{2}$

${~~~~~~~~~~~~~~~}$

CB=√3k

AC=2k

➯Now lets find AB

By pythagoras theorem

AC²=BC²+AB²

(2k)²= BC²+(√3k)²

4k² = BC²+3k²

4k²-3k²=BC²

1k²=BC²

1k=BC

➟sin C cos A - cos C sin A

=$\frac{AB}{AC}~×~\frac{AB}{AC}~-~\frac{BC}{AC}×\frac{BC}{AC}\\ \\=~\frac{1}{2}~×~\frac{1}{2}~-~\frac{√3}{2}~×~\frac{√3}{2}\\\\=~\frac{1}{4}-\frac{3}{4}\\\\\frac{-2}{4}$

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