Question

In ⊿ABC, ∠B = 90 and tan⁡= 3/4 . Find the values of sinA⁡ cosC⁡+ cosA⁡ sinC⁡ .

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{sin\:A\:cos\:C+cos\:A\:sin\:C=1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies \angle B= 90 \degree \\ \\ \tt: \implies tan \: A = \frac{3}{4} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies sin \: A\: cos \: C+ cos \: A \: sin \: C = ?$

• According to given question :

$\tt \circ \: tan \: A = \frac{3}{4} = \frac{p}{b} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {h}^{2} = {3}^{2} + {4}^{2} \\ \\ \tt: \implies {h}^{2} =9 + 16 \\ \\ \tt: \implies {h}^{2} =25 \\ \\ \tt: \implies {h} = \sqrt{25} \\ \\ \green{\tt: \implies {h} =5 } \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies sin \: A \: cos \: C + cos \: A \: sin \: C = \frac{p}{h} \times \frac{b}{h} + \frac{b}{h} \times \frac{p}{h} \\ \\ \tt: \implies sin \: A \: cos \: C + cos \: A \: sin \: C= \frac{3}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{4}{5} \\ \\ \tt: \implies sin \: A \: cos \: C+ cos \: A \: sin \: C = \frac{9}{25} + \frac{16}{25} \\ \\ \tt: \implies sin \: A \: cos \: C + cos \: A \: sin \: C= \frac{9 + 16}{25} \\ \\ \tt: \implies sin \: A\: cos \: C+ cos \: A \: sin \: C= \frac{25}{25} \\ \\ \green{\tt: \implies sin \: A\: cos \: C+ cos \:A \: sin \: C =1}$

Answer 2

Given:

• In ∆ABC , ∠B = 90°
• tanA = 3/4

To find:

• sinA.cosC + cosA.sinC

Solution:

tanA = 3/4

Using Pythagoras theorem

→ Hyp² = 3² + 4²

→ Hyp = √25

→ Hyp = 5

Now,

sinA.cosC + cosA.sinC

• sin∅ = opposite/hypotenuse
• cos∅ = adjacent/hypotenuse

→ 3/4 × 3/4 + 4/5 × 4/5

→ 3²/4² + 4²/5²

→ 9/16 + 16/25

→ 25/25

→ 1 [ Answer ]