In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C.

byAmlesh Mahato •June 08, 2021

0

Solution:

⇨ Given

⇨ tan A = 1/√3 = p/b

⇨ Therefore p = BC = 1 , b = AB = √3, h = AC = ?

⇨ In △ ABC

⇨ From Pythagoras Theorem

$Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics$

$\large \therefore AC^2=AB^2+BC^2$

$\large \Rightarrow AC^2=\left ( \sqrt{3} \right )^2+\left ( 1 \right )^2$

$\large \Rightarrow AC^2=3+1$

$\large \Rightarrow AC^2=4$

$\large \Rightarrow AC^2=2^2$

$\large \Rightarrow AC=2$

(i) sin A cos C + cos A sin C

$\large = \frac{BC}{AC}\times \frac{BC}{AC}+\frac{AB}{AC}\times \frac{AB}{AC}$

$\large = \frac{BC^{2}}{AC^{2}}+\frac{AB^{2}}{AC^{2}}$

$\large = \frac{1^{2}}{2^{2}}+\frac{(\sqrt{3})^{2}}{2^{2}}$

$\large = \frac{1}{4}+\frac{3}{4}$

$\large = \frac{1+3}{4}$

$\large = \frac{4}{4}$

$\large =1$

(ii) cos A cos C – sin A sin C

$In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C.$

$\large =\frac{\sqrt{3}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{\sqrt{3}}{2}$

$\large =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$\large =0$