Question

3 consider the right angled isosceles triangle abc alongside.
a explain why ( widehat{abc}=45^{circ} ).
b use pythagoras theorem to find ab.
c hence find:
i ( sin 45^{circ} ) ii ( cos 45^{circ} ) iii ( \tan 45^{circ} ).
d check your answers using a calculator.

Explanation:

Step1: Explain why \( \angle ABC = 45^{\circ} \)

In a right - angled isosceles triangle \( \triangle ABC \) with \( \angle C=90^{\circ} \) and \( AC = BC=1\ m \).
By the angle - sum property of a triangle \( \angle A+\angle B+\angle C = 180^{\circ} \). Let \( \angle A=\angle B=x \) (since \( AC = BC \), angles opposite to equal sides are equal).
So \( x + x+90^{\circ}=180^{\circ} \), \( 2x=90^{\circ} \), \( x = 45^{\circ} \).

Step2: Use Pythagoras' theorem to find \( AB \)

Pythagoras' theorem states that for a right - triangle \( c^{2}=a^{2}+b^{2} \), where \( c \) is the hypotenuse and \( a,b \) are the legs.
Here \( a = AC = 1\ m \), \( b = BC = 1\ m \) and \( c = AB \).
\( AB^{2}=AC^{2}+BC^{2}\)
\( AB^{2}=1^{2}+1^{2}=1 + 1=2\)
\( AB=\sqrt{2}\ m\)

Step3: Find \( \sin45^{\circ} \)

By the definition of sine in a right - triangle \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
For \( \theta = 45^{\circ} \), opposite side to \( \angle B \) is \( AC = 1\ m \) and hypotenuse \( AB=\sqrt{2}\ m\)
\( \sin45^{\circ}=\frac{AC}{AB}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

Step4: Find \( \cos45^{\circ} \)

By the definition of cosine in a right - triangle \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)
For \( \theta = 45^{\circ} \), adjacent side to \( \angle B \) is \( BC = 1\ m \) and hypotenuse \( AB=\sqrt{2}\ m\)
\( \cos45^{\circ}=\frac{BC}{AB}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

Step5: Find \( \tan45^{\circ} \)

By the definition of tangent in a right - triangle \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)
For \( \theta = 45^{\circ} \), opposite side to \( \angle B \) is \( AC = 1\ m \) and adjacent side to \( \angle B \) is \( BC = 1\ m\)
\( \tan45^{\circ}=\frac{AC}{BC}=\frac{1}{1}=1\)

Answer:

a. In \( \triangle ABC\), \( \angle C = 90^{\circ}\) and \(AC = BC\). By angle - sum property \( \angle A+\angle B+\angle C=180^{\circ}\). Let \( \angle A=\angle B = x\), then \(2x + 90^{\circ}=180^{\circ}\), \(x = 45^{\circ}\), so \( \angle ABC=45^{\circ}\)
b. \(AB=\sqrt{2}\ m\)
c. i. \( \sin45^{\circ}=\frac{\sqrt{2}}{2}\) ii. \( \cos45^{\circ}=\frac{\sqrt{2}}{2}\) iii. \( \tan45^{\circ}=1\)
d. Using a calculator: \( \sin45^{\circ}\approx0.707=\frac{\sqrt{2}}{2}\), \( \cos45^{\circ}\approx0.707=\frac{\sqrt{2}}{2}\), \( \tan45^{\circ}=1\)