Question

In the adjoining figure, find the value of sin B cos C + sin C cos B​

Answer 1

Answer:

$according \: to \: pythagorous \: thereom \\ {ab}^{2} + {ac}^{2} = {bc}^{2} \\ {6}^{2} + {ac}^{2} = {10}^{2} \\ 36 + {ac}^{2} = 100 \\ {ac}^{2} = 100 - 36 \\ {ac}^{2} = 64 \\ ac = \sqrt{64} \\ ac = 8$

$from \: the \: figure \\ \sin(b) = \frac{opposite \: to \: b}{hypotnuse} = \frac{8}{10} \\ \cos(c) = \frac{adjacent \: to \: c}{hypotnuse} = \frac{8}{10} \\ \sin(c) = \frac{opposite \: to \: c}{hypotnuse} = \frac{6}{10} \\ \cos(b) = \frac{adjacent \: to \: b}{hypotnuse} = \frac{6}{10}$

$= \sin(b) \cos(c) + \cos(b) + \sin(c) \\ = \frac{8}{10} \times \frac{8}{10} + \frac{6}{10} \times \frac{6}{10} \\ \frac{64}{100} + \frac{36}{100} \\ = \frac{100}{100} \\ = 1$

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Answer 2

Step-by-step explanation:

Given: Figure attached.

To find: Value of

$sin B cos C + sin C cos B$

Solution:

We know that in a right triangle, one can calculate all the trigonometric ratios for both acute angles.

Step 1 : Find base of ∆ABC

Apply Pythagoras theorem

Hypotenuse² =Base²+ Perpendicular²

$( {10)}^{2} = {Base}^{2} + ( {6)}^{2}$

$AC^2=100-36$

$AC^2=64$

$AC=8$

Step 2: Find trigonometric ratios of expression.

$sin \: B = \frac{ \text{side \: opposite \: to \: angle \: B}}{hypotenuse}$

$sin \: B = \frac{AC}{BC}$

$sin \: B = \frac{8}{10}$

$\bf sin \: B = \frac{4}{5}$

$sin \: C= \frac{ \text{side \: opposite \: to \: angle \: C}}{hypotenuse}$

$sin \: C = \frac{AB}{BC}$

$sin \: C = \frac{6}{10}$

$\bf sin \: C = \frac{3}{5}$

$cos \: B= \frac{ \text{side \: adjacent\: to \: angle \: B}}{hypotenuse}$

$cos\: B = \frac{AB}{BC}$

$cos\: B = \frac{6}{10}$

$\bf cos \: B = \frac{3}{5}$

$cos \: C= \frac{ \text{side \: adjacent\: to \: angle \: C}}{hypotenuse}$

$cos\: C = \frac{AC}{BC}$

$cos\: C = \frac{8}{10}$

$\bf cos \: C = \frac{4}{5}$

Step 3: Put the values of sin C, cos C, cos B and sin C.

$sin B cos C + sin C cos B = \frac{4}{5} \times \frac{4}{5} + \frac{3}{5} \times \frac{3}{5}$

or

$= \frac{16}{25} + \frac{9}{25}$

or

$= \frac{25}{25}$

$sin B cos C + sin C cos B = 1$

option b is correct.

Final answer:

$\bf sin B cos C + sin C cos B = 1$

Option b is correct.

Hope it helps you.

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