Question

If in a triangle ABC is right angled
at B, AB=6 UNITS, BC = 8 units​

Answer 1

Answer:

If AB =6 units and

BC =8 units

By Using Pythagorus formula

(AB)^2 +(BC)^2=(AC)^2

(6)^2+(8)^2=AC^2

36+64=(AC)^2

100=(AC)^2

√100=AC

AC=10 units

Answer 2

Complete question:-

• If in a triangle ABC is right angled at B, AB=6 UNITS, BC = 8 units then find the value of sinA . cosC + cosA . sinC

Given:-

• ABC is a right angle triangle
• ∠ B = 90°
• AB = 6 units
• BC = 8 units

To find:-

• sinA . cosC + cosA . sinC

Solution:-

According to question:-

• sinA = $\sf \dfrac{8}{10}$

• sinC = $\sf \dfrac{6}{10}$

• cosA = $\sf \dfrac{6}{10}$

• cosC = $\sf \dfrac{8}{10}$

$\implies$ sinA . cosC + cosA . sinC

Substituting the values:-

$\implies \sf{ \dfrac{8}{10} \times \dfrac{8}{10} + \dfrac{6}{10} \times \dfrac{6}{10}}$

$\implies \sf{ \dfrac{64}{100} + \dfrac{36}{100}}$

$\implies \sf{ \dfrac{64 + 36}{100}}$

$\implies \sf{ \cancel{ \dfrac{100}{100}}}$

$\implies \bold{\underline{\underline{\boxed{\sf{\red{\dag \; 1}}}}}}$

Therefore, sinA . cosC + cosA . sinC = 1

$\rule{200}{2}$