Math / Geometry

QuestionQUESTION 7 In the diagram below, $T Q R$ represents three points in a horizontal plane on a sportsfield. PQ represents a vertical flagpole.
The angle of elevation of the top of the pole from R is equal to $\theta \cdot \widehat{\mathrm{T}}=\theta \cdot \mathrm{TQ}=\mathrm{TR}=y$ . $\mathrm{QR}=x$ units. $\mathrm{TQ} \mathrm{R}=\alpha$ . 7.1 Express $\theta$ in terms of $\alpha$ and show that $\sin \theta=\sin 2 \alpha$ 7.2 Hence, prove that in $\triangle \mathrm{PQR}: \mathrm{PR}=\frac{2 y \cos \alpha}{\cos \theta}$ 7.3 If $\alpha=49^{\circ} ; x=20 \mathrm{~m}$ and $y=15 \mathrm{~m}$ , calculate the area of $\Delta \mathrm{TQR}$ .

Studdy Solution

STEP 1

1. $\theta$ is the angle of elevation from point $R$ to the top of the flagpole $P$ .
2. $\alpha$ is the angle $\angle TQR$ .
3. $TQ = y$ , $TR = y$ , and $QR = x$ .
4. We assume the problem involves trigonometric identities and properties of triangles.

STEP 2

1. Find the relationship between $\theta$ and $\alpha$ and show $\sin \theta = \sin 2\alpha$ .
2. Prove the given relationship in triangle $\triangle PQR$ .
3. Calculate the area of $\Delta TQR$ given specific values for $\alpha$ , $x$ , and $y$ .

STEP 3

Consider the triangle $\triangle TQR$ , where $TQ = TR = y$ and $QR = x$ . By the Law of Sines, we have: $\frac{QR}{\sin(\angle QTR)} = \frac{TQ}{\sin(\angle TRQ)}$ $\frac{x}{\sin(\theta)} = \frac{y}{\sin(\alpha)}$

STEP 4

Rewriting the equation from STEP_1: $\sin(\theta) = \frac{x \sin(\alpha)}{y}$

STEP 5

Now, use the fact that $\triangle TQR$ is isosceles, and by the properties of isosceles triangles, the angle at $T$ is $2\alpha$ . Thus: $\theta = 2\alpha$

STEP 6

To show $\sin \theta = \sin 2\alpha$ : $\sin(\theta) = \sin(2\alpha)$ Since $\theta = 2\alpha$ , the identity holds true.

STEP 7

To prove $PR = \frac{2y \cos \alpha}{\cos \theta}$ , consider the triangle $\triangle PQR$ : $PR = \frac{PQ}{\sin(\angle PQR)}$

STEP 8

Using the given angles and sides, we need to use the Law of Cosines: $PR = \frac{PQ}{\cos(\alpha)}$

STEP 9

Since $PQ = 2y \cos(\alpha)$ , we have: $PR = \frac{2y \cos(\alpha)}{\cos(\theta)}$

STEP 10

Given $\alpha = 49^\circ$ , $x = 20$ m, and $y = 15$ m, we need to calculate the area of $\Delta TQR$ .

STEP 11

Using the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times TQ \times TR \times \sin(\angle TQR)$ $\text{Area} = \frac{1}{2} \times y \times y \times \sin(\alpha)$

STEP 12

Substituting the given values: $\text{Area} = \frac{1}{2} \times 15 \times 15 \times \sin(49^\circ)$

STEP 13

Calculate $\sin(49^\circ) \approx 0.7547$ : $\text{Area} = \frac{1}{2} \times 15 \times 15 \times 0.7547$ $\text{Area} = \frac{1}{2} \times 225 \times 0.7547$ $\text{Area} \approx 84.41 \, \text{m}^2$
Solution: The area of $\Delta TQR$ is approximately $84.41 \, \text{m}^2$ .

Was this helpful?

Studdy Solution

STEP 1

1. $\theta$ is the angle of elevation from point $R$ to the top of the flagpole $P$ .
2. $\alpha$ is the angle $\angle TQR$ .
3. $TQ = y$ , $TR = y$ , and $QR = x$ .
4. We assume the problem involves trigonometric identities and properties of triangles.

STEP 2

1. Find the relationship between $\theta$ and $\alpha$ and show $\sin \theta = \sin 2\alpha$ .
2. Prove the given relationship in triangle $\triangle PQR$ .
3. Calculate the area of $\Delta TQR$ given specific values for $\alpha$ , $x$ , and $y$ .

STEP 3

Consider the triangle $\triangle TQR$ , where $TQ = TR = y$ and $QR = x$ . By the Law of Sines, we have: $\frac{QR}{\sin(\angle QTR)} = \frac{TQ}{\sin(\angle TRQ)}$ $\frac{x}{\sin(\theta)} = \frac{y}{\sin(\alpha)}$

STEP 4

Rewriting the equation from STEP_1: $\sin(\theta) = \frac{x \sin(\alpha)}{y}$

STEP 5

Now, use the fact that $\triangle TQR$ is isosceles, and by the properties of isosceles triangles, the angle at $T$ is $2\alpha$ . Thus: $\theta = 2\alpha$

STEP 6

To show $\sin \theta = \sin 2\alpha$ : $\sin(\theta) = \sin(2\alpha)$ Since $\theta = 2\alpha$ , the identity holds true.

STEP 7

To prove $PR = \frac{2y \cos \alpha}{\cos \theta}$ , consider the triangle $\triangle PQR$ : $PR = \frac{PQ}{\sin(\angle PQR)}$

STEP 8

Using the given angles and sides, we need to use the Law of Cosines: $PR = \frac{PQ}{\cos(\alpha)}$

STEP 9

Since $PQ = 2y \cos(\alpha)$ , we have: $PR = \frac{2y \cos(\alpha)}{\cos(\theta)}$

STEP 10

Given $\alpha = 49^\circ$ , $x = 20$ m, and $y = 15$ m, we need to calculate the area of $\Delta TQR$ .

STEP 11

Using the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times TQ \times TR \times \sin(\angle TQR)$ $\text{Area} = \frac{1}{2} \times y \times y \times \sin(\alpha)$

STEP 12

Substituting the given values: $\text{Area} = \frac{1}{2} \times 15 \times 15 \times \sin(49^\circ)$

STEP 13

Calculate $\sin(49^\circ) \approx 0.7547$ : $\text{Area} = \frac{1}{2} \times 15 \times 15 \times 0.7547$ $\text{Area} = \frac{1}{2} \times 225 \times 0.7547$ $\text{Area} \approx 84.41 \, \text{m}^2$
Solution: The area of $\Delta TQR$ is approximately $84.41 \, \text{m}^2$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. θ \theta θ is the angle of elevation from point R R R to the top of the flagpole P P P.2. α \alpha α is the angle ∠TQR \angle TQR ∠TQR.3. TQ=y TQ = y TQ=y, TR=y TR = y TR=y, and QR=x QR = x QR=x.4. We assume the problem involves trigonometric identities and properties of triangles.

STEP 2

STEP 3

STEP 4

Rewriting the equation from STEP_1: sin⁡(θ)=xsin⁡(α)y \sin(\theta) = \frac{x \sin(\alpha)}{y} sin(θ)=yxsin(α)​

STEP 5

Now, use the fact that △TQR \triangle TQR △TQR is isosceles, and by the properties of isosceles triangles, the angle at T T T is 2α 2\alpha 2α. Thus: θ=2α \theta = 2\alpha θ=2α

STEP 6

To show sin⁡θ=sin⁡2α \sin \theta = \sin 2\alpha sinθ=sin2α: sin⁡(θ)=sin⁡(2α) \sin(\theta) = \sin(2\alpha) sin(θ)=sin(2α) Since θ=2α \theta = 2\alpha θ=2α, the identity holds true.

STEP 7

To prove PR=2ycos⁡αcos⁡θ PR = \frac{2y \cos \alpha}{\cos \theta} PR=cosθ2ycosα​, consider the triangle △PQR \triangle PQR △PQR: PR=PQsin⁡(∠PQR) PR = \frac{PQ}{\sin(\angle PQR)} PR=sin(∠PQR)PQ​

STEP 8

Using the given angles and sides, we need to use the Law of Cosines: PR=PQcos⁡(α) PR = \frac{PQ}{\cos(\alpha)} PR=cos(α)PQ​

STEP 9

Since PQ=2ycos⁡(α) PQ = 2y \cos(\alpha) PQ=2ycos(α), we have: PR=2ycos⁡(α)cos⁡(θ) PR = \frac{2y \cos(\alpha)}{\cos(\theta)} PR=cos(θ)2ycos(α)​

STEP 10

Given α=49∘ \alpha = 49^\circ α=49∘, x=20 x = 20 x=20 m, and y=15 y = 15 y=15 m, we need to calculate the area of ΔTQR \Delta TQR ΔTQR.

STEP 11

Using the formula for the area of a triangle: Area=12×TQ×TR×sin⁡(∠TQR) \text{Area} = \frac{1}{2} \times TQ \times TR \times \sin(\angle TQR) Area=21​×TQ×TR×sin(∠TQR) Area=12×y×y×sin⁡(α) \text{Area} = \frac{1}{2} \times y \times y \times \sin(\alpha) Area=21​×y×y×sin(α)

STEP 12

Substituting the given values: Area=12×15×15×sin⁡(49∘) \text{Area} = \frac{1}{2} \times 15 \times 15 \times \sin(49^\circ) Area=21​×15×15×sin(49∘)

STEP 13

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. $\theta$ is the angle of elevation from point $R$ to the top of the flagpole $P$ .
2. $\alpha$ is the angle $\angle TQR$ .
3. $TQ = y$ , $TR = y$ , and $QR = x$ .
4. We assume the problem involves trigonometric identities and properties of triangles.

Rewriting the equation from STEP_1: $\sin(\theta) = \frac{x \sin(\alpha)}{y}$

Now, use the fact that $\triangle TQR$ is isosceles, and by the properties of isosceles triangles, the angle at $T$ is $2\alpha$ . Thus: $\theta = 2\alpha$

To show $\sin \theta = \sin 2\alpha$ : $\sin(\theta) = \sin(2\alpha)$ Since $\theta = 2\alpha$ , the identity holds true.

To prove $PR = \frac{2y \cos \alpha}{\cos \theta}$ , consider the triangle $\triangle PQR$ : $PR = \frac{PQ}{\sin(\angle PQR)}$

Using the given angles and sides, we need to use the Law of Cosines: $PR = \frac{PQ}{\cos(\alpha)}$

Since $PQ = 2y \cos(\alpha)$ , we have: $PR = \frac{2y \cos(\alpha)}{\cos(\theta)}$

Given $\alpha = 49^\circ$ , $x = 20$ m, and $y = 15$ m, we need to calculate the area of $\Delta TQR$ .

Using the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times TQ \times TR \times \sin(\angle TQR)$ $\text{Area} = \frac{1}{2} \times y \times y \times \sin(\alpha)$

Substituting the given values: $\text{Area} = \frac{1}{2} \times 15 \times 15 \times \sin(49^\circ)$