In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 3 | Page 44
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.
In ∆RST,
∠S = 90∘, ∠T = 30∘, ∴ ∠R = 60∘
By 30∘ − 60∘ − 90∘ theorem,
\[ \Rightarrow RS = \frac{1}{2} \times 12\]
\[ \Rightarrow RS = 6 cm . . . \left( 1 \right)\]
\[ST = \frac{\sqrt{3}}{2} \times RT\]
\[ \Rightarrow ST = \frac{\sqrt{3}}{2} \times 12\]
\[ \Rightarrow ST = 6\sqrt{3} cm . . . \left( 2 \right)\]
Hence, RS = 6 cm and ST = 6\[\sqrt{3}\] cm.
Explanation:-
Since ∠S = 90°, we can use trigonometric ratios to find the lengths of RS and ST.
Let’s label the sides of triangle RST as follows: RS = a, ST = b, and RT = 12 cm.
Using the trigonometric ratio of sine, we have:
sin(30°) = ST/RT
Substituting the given values, we get:
sin(30°) = b/12
Simplifying, we get:
b = 12 sin(30°) b = 6 cm
Using the Pythagorean theorem, we have:
RS^2 + ST^2 = RT^2
Substituting the given values and the value we just found for b, we get:
a^2 + 6^2 = 12^2
Simplifying, we get:
a^2 + 36 = 144
a^2 = 108
Taking the square root of both sides, we get:
a = sqrt(108)
We can simplify this expression by factoring out the perfect square of 36:
a = 6 sqrt(3)
Therefore, RS = 6 sqrt(3) cm and ST = 6 cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 3 | Page 44