Question

In triangle ABC right angle at B and AB=5 BC=12 find values of sinA secA secC sinC

Answer 1

Answer:

sin A= 0.92 sec A = 1.08 sec C= 2.63 sin C=0.38

Step-by-step explanation:

In a triangle with AB=5 and BC=12 then AC=$\sqrt{AB^2+BC^2 }$ = 13

sin A=BC/AC

sec A = 1/ sin A

sec C=AC/AB

sin C=1/sec C

Answer 2

Answer:

The values are -

= ${\sf{\sin(A) = \dfrac{12}{13}}}$

= $\sf{\sec(A) = \dfrac{13}{5}}$

= $\sf{\sin(C) = \dfrac{5}{13}}$

= $\sf{\sec(C) = \dfrac{13}{12}}$

Step-by-step explanation:

Solution :

In ∆ ABC -

• $\angle$ B = 90°
• AB = 5 units
• BC = 12 units

By Pythagoras theorem,

⇒ (Hypotenuse)² = (Base)² + (Height)²

⇒ (AC)² = (AB)² + (BC)²

⇒ (AC)² = (5)² + (12)²

⇒ (AC)² = 25 + 144

⇒ (AC)² = 169

⇒ AC = 13 units

★ T ratios of $\angle$ A -

• Height = BC = 12
• Base = AB = 5
• Hypotenuse = AC = 13

⇒ $\sf{\sin(A) = \dfrac{BC}{AC}}$

⇒ $\boxed{\sf{\sin(A) = \dfrac{12}{13}}}$

⇒ $\sf{\sec(A) = \dfrac{AC}{AB}}$

⇒ $\boxed{\sf{\sec(A) = \dfrac{13}{5}}}$

$\rule{300}{1.5}$

★ T ratios of $\angle$ C -

• Height = BC = 12
• Base = AB = 5
• Hypotenuse = AC = 13

⇒ $\sf{\sin(C) = \dfrac{AB}{AC}}$

⇒ $\boxed{\sf{\sin(C) = \dfrac{5}{13}}}$

⇒ $\sf{\sec(C) = \dfrac{AC}{BC}}$

⇒ $\boxed{\sf{\sec(C) = \dfrac{13}{12}}}$

Therefore, the values are -

= ${\sf{\sin(A) = \dfrac{12}{13}}}$

= $\sf{\sec(A) = \dfrac{13}{5}}$

= $\sf{\sin(C) = \dfrac{5}{13}}$

= $\sf{\sec(C) = \dfrac{13}{12}}$