In triangle $ABC$, $\angle C=90^\circ$. Point $D$ is the midpoint of $BC$, and $AD=BC$. Find $\sin\angle BAD$.

Step 1: Define variables and relationships

Let's assign a variable to the length of $CD$. Since $D$ is the midpoint of $BC$, $CD = BD$. Therefore, $BC = 2CD$. We are given that $AD = BC$, so $AD = 2x$.

Step 2: Apply Pythagorean theorem in $\triangle ADC$

In the right-angled triangle $ADC$, we can apply the Pythagorean theorem. We have $AC^2 + CD^2 = AD^2$. Substituting the values we defined, we get $AC^2 + x^2 = (2x)^2$. Solving for $AC$, we find $AC = x\sqrt{3}$.

Step 3: Apply Pythagorean theorem in $\triangle ABC$

Next, let's consider the larger right-angled triangle $ABC$. Using the Pythagorean theorem, $AB^2 = AC^2 + BC^2$. Substituting the values we found, $AB^2 = (x\sqrt{3})^2 + (2x)^2$. This simplifies to $AB^2 = 3x^2 + 4x^2 = 7x^2$, so $AB = x\sqrt{7}$.

Step 4: Apply the Law of Cosines in $\triangle ABD$

In triangle $ABD$, we can use the Law of Cosines to find $\cos\angle ABD$. The formula is $AD^2 = AB^2 + BD^2 - 2(AB)(BD)\cos\angle ABD$. Substituting the lengths we found, we get $$(2x)^2 = (x\sqrt{7})^2 + x^2 - 2(x\sqrt{7})(x)\cos\angle ABD$$ . After simplifying, we find that $\cos\angle ABD = \frac{2}{\sqrt{7}}$.

Step 5: Find $\sin\angle ABD$

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, we can find $\sin\angle ABD$. We have $\sin^2\angle ABD = 1 - \left(\frac{2}{\sqrt{7}}\right)^2 = 1 - \frac{4}{7} = \frac{3}{7}$. Taking the square root, $\sin\angle ABD = \frac{\sqrt{3}}{\sqrt{7}} = \frac{\sqrt{21}}{7}$.

Step 6: Apply the Law of Sines in $\triangle ABD$

Finally, we can use the Law of Sines in $\triangle ABD$ to find $\sin\angle BAD$. The formula is $\frac{AD}{\sin\angle ABD} = \frac{BD}{\sin\angle BAD}$. Substituting the known values, we have $\frac{2x}{\frac{\sqrt{21}}{7}} = \frac{x}{\sin\angle BAD}$. Solving for $\sin\angle BAD$, we get $\sin\angle BAD = \frac{\sqrt{21}}{14}$.

More from this chapter

• **EXERCISE 8.3** **1.** Express the trigonometric ratios sin A, tan A and sec A in terms of cot A.
• Cos A / 1 + sin a + 1 + sin a / cos A is equal to 2 sec a
• write all the other trigonometric ratios of angle A in terms of sec A
• Express all the trigonometric ratios in terms of Sin A
• Express the trigonometric ratio sign a sec a tan a in terms of cot a
• Prove that √(sec² A + csc² A) = tan A + cot A
• (secθ - tanθ) = (1)/(m) --- (2) Adding (1) & (2) secθ + tanθ + secθ - tanθ = (1)/(m) + (m)/(1) 2secθ = (1 + m²)/(m) secθ = (1 + m²)/(2m) (Hence proved) 2) If as
• If tanθ + secθ = m, prove that secθ = (m² + 1)/(2m).

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