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# Learn UNIT V: Trigonometry with Free Lessons & Tips

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Answered on 17 Apr

Sadika

Let's denote the height of the hill as hh meters. We can set up a right triangle to represent the situation. From the given information: The angle of elevation of the top of the hill from the foot of the tower is 60 degrees. The angle of elevation of the top of the tower from the foot of the hill is... read more

Let's denote the height of the hill as hh meters. We can set up a right triangle to represent the situation.

From the given information:

1. The angle of elevation of the top of the hill from the foot of the tower is 60 degrees.
2. The angle of elevation of the top of the tower from the foot of the hill is 30 degrees.
3. The height of the tower is 50 meters.

In the right triangle formed by the hill, tower, and the ground, the height of the hill, the height of the tower, and the distance from the foot of the tower to the foot of the hill form the sides of the triangle.

Using trigonometric ratios, we can set up equations based on the given angles and the known side lengths:$For&space;the&space;hill:&space;$&space;\tan(60^\circ)&space;=&space;\frac{h}{x}&space;$&space;For&space;the&space;tower:&space;$&space;\tan(30^\circ)&space;=&space;\frac{h&space;+&space;50}{x}&space;$&space;Where&space;$$&space;x&space;$$&space;is&space;the&space;distance&space;from&space;the&space;foot&space;of&space;the&space;tower&space;to&space;the&space;foot&space;of&space;the&space;hill.&space;We&space;can&space;solve&space;these&space;two&space;equations&space;simultaneously&space;to&space;find&space;the&space;value&space;of&space;$$&space;h&space;$$.&space;First,&space;let's&space;find&space;$$&space;x&space;$$&space;using&space;the&space;fact&space;that&space;the&space;base&space;of&space;both&space;triangles&space;is&space;the&space;same:&space;$&space;x&space;=&space;\frac{h}{\tan(60^\circ)}&space;=&space;\frac{h&space;+&space;50}{\tan(30^\circ)}&space;$&space;$&space;\frac{h}{\frac{\sqrt{3}}{3}}&space;=&space;\frac{h&space;+&space;50}{\frac{1}{\sqrt{3}}}&space;$&space;$&space;\frac{h&space;\times&space;3}{\sqrt{3}}&space;=&space;(h&space;+&space;50)&space;\times&space;\sqrt{3}&space;$&space;$&space;3h&space;=&space;\sqrt{3}(h&space;+&space;50)&space;$&space;$&space;3h&space;=&space;\sqrt{3}h&space;+&space;50\sqrt{3}&space;$&space;$&space;3h&space;-&space;\sqrt{3}h&space;=&space;50\sqrt{3}&space;$&space;$&space;h(3&space;-&space;\sqrt{3})&space;=&space;50\sqrt{3}&space;$&space;$&space;h&space;=&space;\frac{50\sqrt{3}}{3&space;-&space;\sqrt{3}}&space;$$

$Rationalizing&space;the&space;denominator&space;by&space;multiplying&space;both&space;the&space;numerator&space;and&space;the&space;denominator&space;by&space;the&space;conjugate&space;of&space;the&space;denominator:&space;$&space;h&space;=&space;\frac{50\sqrt{3}(3&space;+&space;\sqrt{3})}{(3&space;-&space;\sqrt{3})(3&space;+&space;\sqrt{3})}&space;$&space;$&space;h&space;=&space;\frac{50\sqrt{3}(3&space;+&space;\sqrt{3})}{9&space;-&space;3}&space;$&space;$&space;h&space;=&space;\frac{50\sqrt{3}(3&space;+&space;\sqrt{3})}{6}&space;$&space;$&space;h&space;=&space;\frac{25\sqrt{3}(3&space;+&space;\sqrt{3})}{3}&space;$&space;$&space;h&space;\approx&space;72.4&space;\,&space;\text{meters}$&space;So,&space;the&space;height&space;of&space;the&space;hill&space;is&space;approximately&space;72.4&space;meters.$

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Answered on 17 Apr

Sadika

To find the height of the tower, we can use trigonometry. read more

To find the height of the tower, we can use trigonometry.$TGiven:&space;-&space;The&space;angle&space;of&space;elevation&space;from&space;a&space;point&space;20&space;meters&space;away&space;from&space;the&space;foot&space;of&space;the&space;tower&space;is&space;$$30^\circ$$.&space;Let's&space;denote&space;the&space;height&space;of&space;the&space;tower&space;as&space;$$h$$&space;meters.&space;Using&space;trigonometric&space;ratios,&space;we&space;can&space;set&space;up&space;an&space;equation:&space;$&space;\tan(30^\circ)&space;=&space;\frac{h}{20}&space;$&space;$&space;\frac{1}{\sqrt{3}}&space;=&space;\frac{h}{20}&space;$&space;To&space;solve&space;for&space;$$h$$,&space;we&space;multiply&space;both&space;sides&space;by&space;20:&space;$&space;h&space;=&space;20&space;\times&space;\frac{1}{\sqrt{3}}&space;$&space;$&space;h&space;=&space;\frac{20}{\sqrt{3}}&space;$&space;To&space;rationalize&space;the&space;denominator,&space;we&space;multiply&space;both&space;the&space;numerator&space;and&space;denominator&space;by&space;$$\sqrt{3}$$:&space;$&space;h&space;=&space;\frac{20\sqrt{3}}{3}&space;$&space;So,&space;the&space;height&space;of&space;the&space;tower&space;is&space;approximately&space;$$&space;\frac{20\sqrt{3}}{3}&space;$$&space;meters.$

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Answered on 17 Apr

Sadika

Given: The angles of depression from the top and bottom of the 8 m tall building are 30° and 45°, respectively. We can form two right triangles to represent the situation. For the top of the 8 m tall building: tan⁡(30∘)=8dtan(30∘)=d8 For the bottom of the 8 m tall building: tan⁡(45∘)=8dtan(45∘)=d8 We... read more

Given:

1. The angles of depression from the top and bottom of the 8 m tall building are 30° and 45°, respectively.

We can form two right triangles to represent the situation.

For the top of the 8 m tall building: tan⁡(30∘)=8dtan(30)=d8

For the bottom of the 8 m tall building: tan⁡(45∘)=8dtan(45)=d8

We can solve these two equations simultaneously to find the values of hh and dd.

First, let's solve for dd using either equation (they are the same):

tan⁡(30∘)=8dtan(30)=d8

13=8d3

1=d8

d=83d=83

Now, using the value of dd, we can find the height of the multi-storeyed building using either equation:

tan⁡(30∘)=hdtan(30)=dh

13=h833

1=83

h

h=8h=8

So, the height of the multi-storeyed building is 8 meters, and the distance between the two buildings is 8383

meters.

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Answered on 17 Apr

Sadika

Let's denote: h as the height of the tower. x as the horizontal distance between the tower and the initial position of the car. v as the speed of the car. Given: read more

Let's denote:

• h as the height of the tower.
• x as the horizontal distance between the tower and the initial position of the car.
• v as the speed of the car.

Given:$Let's&space;first&space;find&space;$$&space;x&space;$$&space;using&space;the&space;initial&space;observation:&space;$&space;\tan(30^\circ)&space;=&space;\frac{h}{x}&space;$&space;$&space;\frac{1}{\sqrt{3}}&space;=&space;\frac{h}{x}&space;$&space;$&space;x&space;=&space;h\sqrt{3}&space;$&space;After&space;6&space;seconds,&space;the&space;car&space;travels&space;a&space;horizontal&space;distance&space;of&space;$$&space;v&space;\times&space;6&space;$$&space;meters.&space;So,&space;the&space;new&space;distance&space;from&space;the&space;tower's&space;foot&space;is&space;$$&space;x&space;-&space;v&space;\times&space;6&space;$$.&space;Now,&space;let's&space;find&space;$$&space;x&space;-&space;v&space;\times&space;6&space;$$&space;using&space;the&space;observation&space;after&space;6&space;seconds:&space;$&space;\tan(60^\circ)&space;=&space;\frac{h}{x&space;-&space;v&space;\times&space;6}&space;$&space;$&space;\sqrt{3}&space;=&space;\frac{h}{x&space;-&space;v&space;\times&space;6}&space;$&space;$&space;x&space;-&space;v&space;\times&space;6&space;=&space;\frac{h}{\sqrt{3}}&space;$&space;Since&space;$$&space;x&space;=&space;h\sqrt{3}&space;$$,&space;we&space;can&space;substitute&space;$$&space;x&space;$$&space;in&space;the&space;equation:$

$$&space;h\sqrt{3}&space;-&space;v&space;\times&space;6&space;=&space;\frac{h}{\sqrt{3}}&space;$&space;$&space;h\sqrt{3}&space;-&space;\frac{h}{\sqrt{3}}&space;=&space;v&space;\times&space;6&space;$&space;$&space;h(\sqrt{3}&space;-&space;\frac{1}{\sqrt{3}})&space;=&space;v&space;\times&space;6&space;$&space;$&space;h(\frac{3&space;-&space;1}{\sqrt{3}})&space;=&space;v&space;\times&space;6&space;$&space;$&space;h\frac{2}{\sqrt{3}}&space;=&space;v&space;\times&space;6&space;$&space;$&space;h\frac{2\sqrt{3}}{3}&space;=&space;v&space;\times&space;6&space;$&space;$&space;v&space;=&space;\frac{h\sqrt{3}}{3}&space;$&space;Since&space;we&space;know&space;$$&space;v&space;$$&space;and&space;$$&space;h&space;$$&space;(height&space;of&space;the&space;tower),&space;we&space;can&space;find&space;the&space;time&space;taken&space;by&space;the&space;car&space;to&space;reach&space;the&space;foot&space;of&space;the&space;tower&space;from&space;the&space;initial&space;observation&space;point.&space;Since&space;the&space;car&space;travels&space;a&space;distance&space;of&space;$$&space;h\sqrt{3}&space;$$&space;meters&space;in&space;$$&space;6&space;$$&space;seconds,&space;its&space;speed&space;$$&space;v&space;$$&space;is&space;$$&space;\frac{h\sqrt{3}}{3}&space;$$&space;meters&space;per&space;second.&space;The&space;time&space;taken&space;by&space;the&space;car&space;to&space;travel&space;the&space;distance&space;$$&space;h\sqrt{3}&space;$$&space;meters&space;is:&space;$&space;\text{Time}&space;=&space;\frac{\text{Distance}}{\text{Speed}}&space;=&space;\frac{h\sqrt{3}}{\frac{h\sqrt{3}}{3}}&space;=&space;3&space;\text{&space;seconds}&space;$&space;So,&space;the&space;time&space;taken&space;by&space;the&space;car&space;to&space;reach&space;the&space;foot&space;of&space;the&space;tower&space;from&space;the&space;initial&space;observation&space;point&space;is&space;3&space;seconds.$

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Answered on 17 Apr

Sadika

Given: We need to compute the value of: read more

Given: $$&space;\sec(\theta)&space;+&space;\tan(\theta)&space;=&space;7&space;$$

We need to compute the value of:
$$&space;\sec(\theta)&space;-&space;\tan(\theta)&space;$&space;We&space;can&space;solve&space;this&space;problem&space;by&space;manipulating&space;the&space;given&space;equation.&space;We&space;know&space;that:&space;$&space;\sec(\theta)&space;=&space;\frac{1}{\cos(\theta)}&space;$&space;$&space;\tan(\theta)&space;=&space;\frac{\sin(\theta)}{\cos(\theta)}&space;$&space;So,&space;let's&space;rewrite&space;the&space;given&space;equation&space;with&space;these&space;substitutions:&space;$&space;\frac{1}{\cos(\theta)}&space;+&space;\frac{\sin(\theta)}{\cos(\theta)}&space;=&space;7&space;$$

$Now,&space;let's&space;combine&space;the&space;fractions:&space;$&space;\frac{1&space;+&space;\sin(\theta)}{\cos(\theta)}&space;=&space;7&space;$&space;Multiply&space;both&space;sides&space;by&space;$$&space;\cos(\theta)&space;$$:&space;$&space;1&space;+&space;\sin(\theta)&space;=&space;7&space;\cos(\theta)&space;$$

$Now,&space;let's&space;square&space;both&space;sides&space;to&space;eliminate&space;the&space;square&space;root:&space;$&space;(1&space;+&space;\sin(\theta))^2&space;=&space;(7&space;\cos(\theta))^2&space;$&space;$&space;1&space;+&space;2\sin(\theta)&space;+&space;\sin^2(\theta)&space;=&space;49&space;\cos^2(\theta)&space;$&space;$&space;1&space;+&space;2\sin(\theta)&space;+&space;\sin^2(\theta)&space;=&space;49(1&space;-&space;\sin^2(\theta))&space;$&space;$&space;1&space;+&space;2\sin(\theta)&space;+&space;\sin^2(\theta)&space;=&space;49&space;-&space;49\sin^2(\theta)&space;$&space;$&space;2\sin(\theta)&space;+&space;\sin^2(\theta)&space;+&space;49\sin^2(\theta)&space;=&space;49&space;-&space;1&space;$&space;$&space;50\sin^2(\theta)&space;+&space;2\sin(\theta)&space;-&space;48&space;=&space;0&space;$$

$Now,&space;this&space;is&space;a&space;quadratic&space;equation&space;in&space;terms&space;of&space;$$&space;\sin(\theta)&space;$$.&space;Let's&space;solve&space;it&space;using&space;the&space;quadratic&space;formula:&space;$&space;\sin(\theta)&space;=&space;\frac{-b&space;\pm&space;\sqrt{b^2&space;-&space;4ac}}{2a}&space;$&space;where&space;$$&space;a&space;=&space;50&space;$$,&space;$$&space;b&space;=&space;2&space;$$,&space;and&space;$$&space;c&space;=&space;-48&space;$$.&space;$&space;\sin(\theta)&space;=&space;\frac{-2&space;\pm&space;\sqrt{2^2&space;-&space;4&space;\cdot&space;50&space;\cdot&space;(-48)}}{2&space;\cdot&space;50}&space;$&space;$&space;\sin(\theta)&space;=&space;\frac{-2&space;\pm&space;\sqrt{4&space;+&space;9600}}{100}&space;$&space;$&space;\sin(\theta)&space;=&space;\frac{-2&space;\pm&space;\sqrt{9604}}{100}&space;$&space;$&space;\sin(\theta)&space;=&space;\frac{-2&space;\pm&space;98}{100}&space;$$

$Now,&space;we&space;have&space;two&space;possible&space;values&space;for&space;$$&space;\sin(\theta)&space;$$:&space;1.&space;$$&space;\sin(\theta)&space;=&space;\frac{-2&space;+&space;98}{100}&space;=&space;\frac{96}{100}&space;=&space;0.96&space;$$&space;2.&space;$$&space;\sin(\theta)&space;=&space;\frac{-2&space;-&space;98}{100}&space;=&space;\frac{-100}{100}&space;=&space;-1&space;$$&space;However,&space;$$&space;\sin(\theta)&space;$$&space;must&space;be&space;between&space;-1&space;and&space;1,&space;so&space;the&space;second&space;solution&space;is&space;extraneous.&space;Therefore,&space;$$&space;\sin(\theta)&space;=&space;0.96&space;$$.&space;Now,&space;we&space;can&space;find&space;$$&space;\cos(\theta)&space;$$&space;using&space;the&space;fact&space;that&space;$$&space;\sin^2(\theta)&space;+&space;\cos^2(\theta)&space;=&space;1&space;$$:&space;$&space;\cos^2(\theta)&space;=&space;1&space;-&space;\sin^2(\theta)&space;$&space;$&space;\cos^2(\theta)&space;=&space;1&space;-&space;0.96^2&space;$&space;$&space;\cos^2(\theta)&space;=&space;1&space;-&space;0.9216&space;$&space;$&space;\cos^2(\theta)&space;=&space;0.0784&space;$&space;$&space;\cos(\theta)&space;=&space;\pm&space;\sqrt{0.0784}&space;$&space;$&space;\cos(\theta)&space;=&space;\pm&space;0.28&space;$$

$Since&space;$$&space;\sec(\theta)&space;=&space;\frac{1}{\cos(\theta)}&space;$$&space;and&space;$$&space;\tan(\theta)&space;=&space;\frac{\sin(\theta)}{\cos(\theta)}&space;$$,&space;we&space;can&space;compute&space;them:&space;$&space;\sec(\theta)&space;=&space;\frac{1}{0.28}&space;=&space;\frac{100}{28}&space;=&space;\frac{25}{7}&space;$&space;$&space;\tan(\theta)&space;=&space;\frac{0.96}{0.28}&space;\approx&space;3.428&space;$&space;Finally,&space;we&space;can&space;find&space;$$&space;\sec(\theta)&space;-&space;\tan(\theta)&space;$$:&space;$&space;\sec(\theta)&space;-&space;\tan(\theta)&space;=&space;\frac{25}{7}&space;-&space;3.428&space;=&space;\frac{25}{7}&space;-&space;\frac{240}{70}&space;=&space;\frac{25}{7}&space;-&space;\frac{120}{35}&space;=&space;\frac{25&space;\times&space;10&space;-&space;120}{7&space;\times&space;10}&space;=&space;\frac{130&space;-&space;120}{70}&space;=&space;\frac{10}{70}&space;=&space;\frac{1}{7}&space;$&space;So,&space;$$&space;\sec(\theta)&space;-&space;\tan(\theta)&space;=&space;\frac{1}{7}&space;$$.$

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Answered on 17 Apr

Sadika

Given:tan⁡(θ)+cot⁡(θ)=5 We need to find the value of: read more

Given:tan⁡(θ)+cot⁡(θ)=5

We need to find the value of:
$$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;$&space;We&space;know&space;that:&space;$&space;\cot(\theta)&space;=&space;\frac{1}{\tan(\theta)}&space;$&space;Let's&space;rewrite&space;the&space;given&space;equation&space;in&space;terms&space;of&space;only&space;$$&space;\tan(\theta)&space;$$:&space;$&space;\tan(\theta)&space;+&space;\frac{1}{\tan(\theta)}&space;=&space;5&space;$&space;Multiplying&space;both&space;sides&space;by&space;$$&space;\tan(\theta)&space;$$:&space;$&space;\tan^2(\theta)&space;+&space;1&space;=&space;5\tan(\theta)&space;$$

$Now,&space;we&space;need&space;to&space;find&space;$$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;$$,&space;which&space;can&space;be&space;written&space;as:&space;$&space;(\tan^2(\theta)&space;+&space;1)&space;+&space;(\cot^2(\theta)&space;-&space;1)&space;$&space;$&space;(\tan^2(\theta)&space;+&space;\cot^2(\theta))&space;+&space;(1&space;-&space;1)&space;$&space;$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;$&space;Now,&space;we&space;already&space;have&space;$$&space;\tan^2(\theta)&space;+&space;1&space;$$&space;from&space;the&space;given&space;equation.&space;Let's&space;find&space;$$&space;\cot^2(\theta)&space;-&space;1&space;$$:&space;$&space;\cot(\theta)&space;=&space;\frac{1}{\tan(\theta)}&space;$&space;$&space;\cot^2(\theta)&space;=&space;\frac{1}{\tan^2(\theta)}&space;$&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{1}{\tan^2(\theta)}&space;-&space;1&space;$&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{1&space;-&space;\tan^2(\theta)}{\tan^2(\theta)}&space;$&space;We&space;know&space;that&space;$$&space;\tan^2(\theta)&space;+&space;1&space;=&space;5\tan(\theta)&space;$$,&space;so&space;$$&space;1&space;-&space;\tan^2(\theta)&space;=&space;-\tan^2(\theta)&space;+&space;1&space;$$:&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{-\tan^2(\theta)&space;+&space;1}{\tan^2(\theta)}&space;$&space;Now,&space;we&space;can&space;substitute&space;the&space;value&space;of&space;$$&space;1&space;-&space;\tan^2(\theta)&space;$$&space;from&space;the&space;given&space;equation:&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{-(\tan^2(\theta)&space;+&space;1)&space;+&space;1}{\tan^2(\theta)}&space;$&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{-5\tan(\theta)&space;+&space;1&space;+&space;1}{\tan^2(\theta)}&space;$&space;$&space;\cot^2(\theta)&space;-&space;1&space;=&space;\frac{-5\tan(\theta)&space;+&space;2}{\tan^2(\theta)}&space;$$

$Now,&space;let's&space;add&space;$$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;$$:&space;$&space;(\tan^2(\theta)&space;+&space;1)&space;+&space;(\cot^2(\theta)&space;-&space;1)&space;$&space;$&space;\tan^2(\theta)&space;+&space;1&space;-&space;1&space;+&space;\cot^2(\theta)&space;$&space;$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;$&space;$&space;(5\tan(\theta))&space;+&space;\frac{-5\tan(\theta)&space;+&space;2}{\tan^2(\theta)}&space;$&space;$&space;\frac{(5\tan^3(\theta))&space;+&space;(-5\tan(\theta)&space;+&space;2)}{\tan^2(\theta)}&space;$&space;$&space;\frac{5\tan^3(\theta)&space;-&space;5\tan(\theta)&space;+&space;2}{\tan^2(\theta)}&space;$&space;$&space;\frac{5(\tan^3(\theta)&space;-&space;\tan(\theta))&space;+&space;2}{\tan^2(\theta)}&space;$&space;So,&space;$$&space;\tan^2(\theta)&space;+&space;\cot^2(\theta)&space;=&space;\frac{5(\tan^3(\theta)&space;-&space;\tan(\theta))&space;+&space;2}{\tan^2(\theta)}&space;$$.$

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Answered on 17 Apr

Sadika

As the angle increases from 0° to 90°, the value of the cosine function will decrease. Here's why: At 0∘0∘, the cosine function is at its maximum value of 1. As the angle increases from 0∘0∘ to 90∘90∘, the cosine function gradually decreases. At 90∘90∘, the cosine function reaches its minimum... read more

As the angle increases from 0° to 90°, the value of the cosine function will decrease.

Here's why:

1. At 0∘0, the cosine function is at its maximum value of 1.
2. As the angle increases from 0∘0 to 90∘90, the cosine function gradually decreases.
3. At 90∘90, the cosine function reaches its minimum value of 0.

This behavior is evident if you look at the unit circle representation of trigonometric functions. As you move counterclockwise from the initial point (1, 0) on the unit circle (corresponding to 0∘0), the x-coordinate (which represents the cosine function) decreases until it reaches 0 at 90∘90.

So, as the value of the cosine function increases from 0∘0 to 90∘90, its actual value decreases from 1 to 0.

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Answered on 17 Apr

Sadika

Given: We need to find all other values of trigonometric ratios based on this information. read more

Given:
$$&space;\csc(\theta)&space;=&space;\frac{4}{3}&space;$$

We need to find all other values of trigonometric ratios based on this information.

$We&space;know&space;that:&space;$&space;\csc(\theta)&space;=&space;\frac{1}{\sin(\theta)}&space;$&space;So,&space;if&space;$$&space;\csc(\theta)&space;=&space;\frac{4}{3}&space;$$,&space;then:&space;$&space;\sin(\theta)&space;=&space;\frac{1}{\csc(\theta)}&space;=&space;\frac{1}{\frac{4}{3}}&space;=&space;\frac{3}{4}&space;$&space;Now,&space;we&space;can&space;find&space;other&space;trigonometric&space;ratios&space;using&space;the&space;value&space;of&space;$$&space;\sin(\theta)&space;$$.&space;$&space;\sin(\theta)&space;=&space;\frac{3}{4}&space;$$

$Using&space;the&space;Pythagorean&space;identity&space;$$&space;\sin^2(\theta)&space;+&space;\cos^2(\theta)&space;=&space;1&space;$$,&space;we&space;can&space;find&space;$$&space;\cos(\theta)&space;$$:&space;$&space;\cos^2(\theta)&space;=&space;1&space;-&space;\sin^2(\theta)&space;=&space;1&space;-&space;\left(\frac{3}{4}\right)^2&space;=&space;1&space;-&space;\frac{9}{16}&space;=&space;\frac{16}{16}&space;-&space;\frac{9}{16}&space;=&space;\frac{7}{16}&space;$&space;$&space;\cos(\theta)&space;=&space;\pm&space;\sqrt{\frac{7}{16}}&space;=&space;\pm&space;\frac{\sqrt{7}}{4}&space;$&space;Since&space;$$&space;\theta&space;$$&space;is&space;in&space;the&space;first&space;quadrant&space;(since&space;$$&space;\csc(\theta)&space;$$&space;is&space;positive),&space;$$&space;\cos(\theta)&space;$$&space;is&space;positive.&space;Therefore:&space;$&space;\cos(\theta)&space;=&space;\frac{\sqrt{7}}{4}&space;$$

$Using&space;the&space;fact&space;that&space;$$&space;\tan(\theta)&space;=&space;\frac{\sin(\theta)}{\cos(\theta)}&space;$$,&space;we&space;can&space;find&space;$$&space;\tan(\theta)&space;$$:&space;$&space;\tan(\theta)&space;=&space;\frac{\sin(\theta)}{\cos(\theta)}&space;=&space;\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}&space;=&space;\frac{3}{\sqrt{7}}&space;=&space;\frac{3\sqrt{7}}{7}&space;$&space;Now,&space;we&space;can&space;find&space;other&space;trigonometric&space;ratios&space;using&space;the&space;previously&space;found&space;values:&space;$&space;\cot(\theta)&space;=&space;\frac{1}{\tan(\theta)}&space;=&space;\frac{7}{3\sqrt{7}}&space;=&space;\frac{7\sqrt{7}}{21}&space;$&space;$&space;\sec(\theta)&space;=&space;\frac{1}{\cos(\theta)}&space;=&space;\frac{1}{\frac{\sqrt{7}}{4}}&space;=&space;\frac{4}{\sqrt{7}}&space;=&space;\frac{4\sqrt{7}}{7}&space;$&space;$&space;\cot(\theta)&space;=&space;\frac{1}{\tan(\theta)}&space;=&space;\frac{1}{\frac{3\sqrt{7}}{7}}&space;=&space;\frac{7}{3\sqrt{7}}&space;=&space;\frac{7\sqrt{7}}{21}&space;$$

$So,&space;the&space;other&space;values&space;of&space;trigonometric&space;ratios&space;are:&space;$&space;\sin(\theta)&space;=&space;\frac{3}{4}&space;$&space;$&space;\cos(\theta)&space;=&space;\frac{\sqrt{7}}{4}&space;$&space;$&space;\tan(\theta)&space;=&space;\frac{3\sqrt{7}}{7}&space;$&space;$&space;\cot(\theta)&space;=&space;\frac{7\sqrt{7}}{21}&space;$&space;$&space;\sec(\theta)&space;=&space;\frac{4\sqrt{7}}{7}&space;$&space;$&space;\cot(\theta)&space;=&space;\frac{7\sqrt{7}}{21}&space;$$

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Answered on 17 Apr

Sadika

To find the value of read more

To find the value of$T&space;$$&space;\sin(45^\circ)&space;-&space;\cos(45^\circ)&space;$$,&space;we&space;first&space;need&space;to&space;determine&space;the&space;values&space;of&space;$$&space;\sin(45^\circ)&space;$$&space;and&space;$$&space;\cos(45^\circ)&space;$$.&space;Given&space;that&space;$$&space;45^\circ&space;$$&space;is&space;one&space;of&space;the&space;angles&space;for&space;which&space;the&space;values&space;of&space;sine&space;and&space;cosine&space;are&space;known,&space;we&space;have:&space;$&space;\sin(45^\circ)&space;=&space;\cos(45^\circ)&space;=&space;\frac{\sqrt{2}}{2}&space;$&space;Now,&space;we&space;can&space;substitute&space;these&space;values&space;into&space;the&space;expression:&space;$&space;\sin(45^\circ)&space;-&space;\cos(45^\circ)&space;=&space;\frac{\sqrt{2}}{2}&space;-&space;\frac{\sqrt{2}}{2}&space;$&space;$&space;\sin(45^\circ)&space;-&space;\cos(45^\circ)&space;=&space;0&space;$&space;So,&space;$$&space;\sin(45^\circ)&space;-&space;\cos(45^\circ)&space;=&space;0&space;$$.$

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Answered on 17 Apr

Sadika

Given: We need to prove that read more

Given: $$&space;\sin(B)&space;=&space;\sin(Q)&space;$$

We need to prove that $$$&space;\angle&space;B&space;=&space;\angle&space;Q&space;$$.&space;In&space;trigonometry,&space;the&space;sine&space;function&space;is&space;positive&space;in&space;the&space;first&space;and&space;second&space;quadrants.&space;Since&space;$$&space;B&space;$$&space;and&space;$$&space;Q&space;$$&space;are&space;acute&space;angles,&space;they&space;lie&space;in&space;the&space;first&space;quadrant.&space;In&space;the&space;first&space;quadrant,&space;both&space;$$&space;\sin(B)&space;$$&space;and&space;$$&space;\sin(Q)&space;$$&space;are&space;positive,&space;so&space;we&space;can&space;conclude&space;that&space;$$&space;B&space;$$&space;and&space;$$&space;Q&space;$$&space;are&space;acute&space;angles&space;less&space;than&space;$$&space;90^\circ&space;$$.$
$Now,&space;we&space;know&space;that&space;the&space;sine&space;function&space;is&space;a&space;periodic&space;function&space;with&space;a&space;period&space;of&space;$$&space;360^\circ&space;$$.&space;This&space;means&space;that&space;if&space;two&space;angles&space;have&space;the&space;same&space;sine&space;value,&space;they&space;could&space;be&space;coterminal.&space;However,&space;since&space;both&space;$$&space;B&space;$$&space;and&space;$$&space;Q&space;$$&space;are&space;acute&space;angles,&space;they&space;must&space;lie&space;between&space;$$&space;0^\circ&space;$$&space;and&space;$$&space;90^\circ&space;$$.&space;In&space;this&space;range,&space;the&space;sine&space;function&space;is&space;strictly&space;increasing.&space;Therefore,&space;if&space;two&space;acute&space;angles&space;have&space;the&space;same&space;sine&space;value,&space;they&space;must&space;be&space;the&space;same&space;angle.&space;Hence,&space;we&space;can&space;conclude&space;that&space;$$&space;\angle&space;B&space;=&space;\angle&space;Q&space;$$.$

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