{"id":327,"date":"2017-07-19T02:42:51","date_gmt":"2017-07-18T21:12:51","guid":{"rendered":"http:\/\/ibelitetutor.com\/blog\/?p=327"},"modified":"2024-05-22T15:01:53","modified_gmt":"2024-05-22T09:31:53","slug":"solve-basic-problems-trigonometry","status":"publish","type":"post","link":"http:\/\/ibelitetutor.com\/blog\/solve-basic-problems-trigonometry\/","title":{"rendered":"How to solve basic problems in trigonometry?(concept-1)"},"content":{"rendered":"

Mathematics tutors<\/h2>\n

Our IB Maths Tutors<\/a> say that Trigonometry is a very important and interesting branch of Maths<\/strong><\/p>\n

Trigonometry is one of the branches of Mathematics that fascinates Mathematicians. It tells about the relationships between the angles and sides of a triangle. Word trigonometry was started from the Greek word, where,<\/p>\n

\u2018TRI<\/strong>\u2018 =\u00a0Three<\/em><\/p>\n

\u2018GON<\/strong>\u2018= sides and the\u00a0\u2018METRON\u2019 =<\/strong>\u00a0to\u00a0measure.\u00a0<\/strong>It’s an ancient and probably most widely applicable branch in Mathematics.<\/p>\n

Parts of Trigonometry-<\/h3>\n

For basic learning, I am dividing trigonometry in two-part:-<\/p>\n

1 Trigonometry based on right triangles<\/p>\n

1 Trigonometry based on non-right triangles.<\/p>\n

In this post, I will only discuss Trigonometry based on right triangles.<\/p>\n

In a right triangle, there are three sides hypotenuse (the longest side), adjacent\u00a0side(base) and the opposite side(perpendicular).<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\"<img<\/a><\/p>\n

Pythagoras\u00a0theorem<\/h3>\n

If we know any two sides of the right triangle, we can find the third side by using Pythagoras\u00a0theorem:-<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0P\u00b2+B\u00b2=H\u00b2<\/strong><\/h5>\n

In a right triangle, the ratios of sides can be written in many ways like P\/H, B\/H, P\/B, H\/P, H\/B, B\/P. each ratio is called a trigonometric ratio and these are represented by Latin words Sin, Cos, Tan, Cosec, Sec and Cot.
\nIn a right triangle<\/p>\n

Sin =P\/H, Cos =B\/H, \u00a0Tan= P\/B, Cosec= H\/P, Sec =H\/B, Cot= B\/P,<\/strong><\/p>\n

Mathematics tutors use\u00a0these three concepts to solve trigonometric\u00a0problems<\/p>\n

Concept-1 <\/span><\/p>\n

In a trigonometric problem, If we are given a trigonometric ratio or asked about it, we will first find all three sides of the triangle<\/span><\/strong><\/p>\n

Using the above concept and above formulas, we can solve the first type of trigonometric problems<\/h4>\n

Sample Problems<\/h4>\n

Example-1\u00a0\u00a0<\/b>If sin \u03b8 = 8\/17, find other trigonometric ratios of \u03b8.<\/strong><\/p>\n

Ans-<\/p>\n

Let us draw a \u2206 OMP in which \u2220M = 90\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\"<img<\/a><\/p>\n

 <\/p>\n

Then sin \u03b8 = MP\/OP = 8\/17.<\/p>\n

Let MP = 8k and OP = 17k, where k is positive.<\/p>\n

By Pythagoras\u2019 theorem, we get<\/p>\n

OP2<\/sup>\u00a0= OM2<\/sup>\u00a0+ MP2<\/sup><\/p>\n

\u21d2 OM2<\/sup>\u00a0= OP2<\/sup>\u00a0\u2013 MP2<\/sup><\/p>\n

\u21d2 OM2<\/sup>\u00a0= [(17k)2<\/sup>\u00a0\u2013 (8k)2<\/sup>]<\/p>\n

\u21d2 OM2<\/sup>\u00a0= [289k2<\/sup>\u00a0\u2013 64k2<\/sup>]<\/p>\n

\u21d2 OM2<\/sup>\u00a0= 225k2<\/sup><\/p>\n

\u21d2 OM = \u221a(225k2<\/sup>)<\/p>\n

\u21d2 OM = 15k<\/p>\n

Therefore, sin \u03b8 = MP\/OP = 8k\/17k = 8\/17<\/p>\n

cos \u03b8 = OM\/OP = 15k\/17k = 15\/17<\/strong><\/p>\n

tan \u03b8 = Sin \u03b8\/Cos \u03b8 = (8\/17 \u00d7 17\/15) = 8\/15<\/strong><\/p>\n

csc \u03b8 = 1\/sin \u03b8 = 17\/8<\/strong><\/p>\n

sec \u03b8 = 1\/cos \u03b8 = 17\/15 and<\/strong><\/p>\n

cot \u03b8 = 1\/tan \u03b8 = 15\/8.<\/strong><\/p>\n

Example-2\u00a0If cos\u00a0\u03b8 = 3\/5, find the value of (5cosec \u03b8 – 4 tan \u03b8)\/(sec \u03b8 + cot \u03b8)<\/strong><\/p>\n

Answer-<\/strong><\/p>\n

Let us draw an \u2206 ABC in which \u2220B = 90\u00b0. \u00a0\"<img<\/a><\/p>\n

Let \u2220A = \u03b8\u00b0<\/p>\n

Then cos \u03b8 = AB\/AC = 3\/5.<\/p>\n

Let AB = 3k and AC = 5k, where k is positive.<\/p>\n

By Pythagoras\u2019 theorem, we get<\/p>\n

AC2<\/sup>\u00a0= AB2<\/sup>\u00a0+ BC2<\/sup><\/p>\n

\u21d2 BC2<\/sup>\u00a0= AC2<\/sup>\u00a0\u2013 AB2<\/sup><\/p>\n

\u21d2 BC2<\/sup>\u00a0= [(5k)2<\/sup>\u00a0\u2013 (3k)2<\/sup>]<\/p>\n

\u21d2 BC2<\/sup>\u00a0= [25k2<\/sup>\u00a0\u2013 9k2<\/sup>]<\/p>\n

\u21d2 BC2<\/sup>\u00a0= 16k2<\/sup><\/p>\n

\u21d2 BC = \u221a(16k2<\/sup>)<\/p>\n

\u21d2 BC = 4k<\/p>\n

Therefore, sec \u03b8 = 1\/cos \u03b8 = 5\/3<\/p>\n

tan \u03b8 = BC\/AB =4k\/3k = 4\/3<\/p>\n

cot \u03b8 = 1\/tan \u03b8 = 3\/4 and<\/p>\n

csc \u03b8 = AC\/BC = 5k\/4k = 5\/4<\/p>\n

Now (5cosec \u03b8 -4 tan \u03b8)\/(sec \u03b8 + cot \u03b8)<\/strong><\/p>\n

= (5 \u00d7 5\/4 – 4 \u00d7 4\/3)\/(5\/3 + 3\/4)<\/p>\n

= (25\/4 -16\/3)\/(5\/3 +3\/4)<\/p>\n

= 11\/12 \u00d7 12\/29<\/p>\n

= 11\/29<\/p>\n

In next post of Mathematics tutors series, we will discuss the remaining two concepts of non-right triangle trigonometry<\/p>\n

To get a better understanding of the concept you should check all my posts of Mathematics tutors series. you can check them by clicking on the links given below<\/p>\n

First post- concept one\u00a0(current\u00a0post)<\/span><\/h4>\n

Second post- concept two(current\u00a0post)<\/span><\/h4>\n

Third post -Concept three<\/span><\/h4>\n

Fourth post -Concept four<\/span><\/h4>\n

\"ib<\/p>\n\n

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    <\/ul><\/div>\n
    \n
    \n<\/fieldset>\n

    \n<\/p>

    <\/div>\n<\/form>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"

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